51. ∫ sec 6 x d x \int \sec^6x dx ∫ sec 6 x d x
∫ sec 6 x d x = ∫ sec 2 x s e c 4 x d x ( D ( t a n x ) → s e c 2 x ) = s e c 4 x t a n x − ∫ 4 s e c 3 x s e c x t a n x t a n x d x = s e c 4 x t a n x − 4 ∫ s e c 4 x t a n 2 x d x = s e c 4 x t a n x − 4 ∫ s e c 4 x s e c 2 x − s e c 4 x d x = s e c 4 x t a n x − 4 ∫ s e c 6 x d x + 4 ∫ s e c 4 x d x 5 ∫ s e c 6 x d x = s e c 4 x t a n x + 4 ∫ s e c 4 d x ∫ s e c 4 d x = ∫ sec 2 x ( 1 + tan 2 x ) d x = ∫ s e c 2 x d x + ∫ s e c 2 x t a n 2 x d x = tan x + ∫ s e c 2 x t a n 2 x d x ∫ s e c 2 x t a n 2 x d x ( u = t a n x , d u = s e c 2 x d x ) = ∫ u 2 d u = 1 3 u 3 = 1 3 tan 3 x 5 ∫ s e c 6 x d x = s e c 4 x t a n x + 4 ∫ s e c 4 d x = s e c 4 x t a n x + 4 ( t a n x + 1 3 tan 3 x ) ∴ ∫ s e c 6 x d x = 1 5 s e c 4 x t a n x + 4 5 t a n x + 4 15 t a n 3 x + C = 1 5 ( 1 + 2 t a n 2 x + t a n 4 x ) t a n x + 4 5 t a n x + 4 15 t a n 3 x + C = 1 5 t a n 5 x + 5 5 t a n x + 10 15 t a n 3 x + C = 1 5 t a n 5 x + 2 3 t a n 3 x + t a n x + C
\begin{aligned}
&\int \sec^6x dx =\int \sec^2x sec^4 x dx\\
&(D( tanx ) \rightarrow sec^2x) \\
&=sec^4x tanx-\int 4sec^3xsecxtanxtanxdx\\
&=sec^4xtanx-4\int sec^4xtan^2x dx\\
&=sec^4xtanx-4\int sec^4xsec^2x-sec^4x dx\\
&=sec^4xtanx-4\int sec^6x dx+4\int sec^4x dx\\
&5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\
&\int sec^4 dx=\int \sec^2x (1+\tan^2x) dx=\int sec^2x dx+\int sec^2xtan^2xdx\\
&=\tan{x}+\int sec^2xtan^2x dx\\
&\int sec^2xtan^2x dx (u=tan x, du= sec^2x dx)\\
&=\int u^2 du = \frac{1}{3}u^3=\frac{1}{3}\tan^3{x}\\
&5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\
&=sec^4xtanx+4(tan x+\frac{1}{3}\tan^3{x})\\
&\therefore \int sec^6x dx=\frac{1}{5}sec^4xtanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\
&=\frac{1}{5}(1+2tan^2x+tan^4x)tanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\
&=\frac{1}{5}tan^5x+\frac{5}{5}tanx+\frac{10}{15}tan^3x +C\\
&=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\
\end{aligned}
∫ sec 6 x d x = ∫ sec 2 x s e c 4 x d x ( D ( t a n x ) → s e c 2 x ) = s e c 4 x t a n x − ∫ 4 s e c 3 x s e c x t a n x t a n x d x = s e c 4 x t a n x − 4 ∫ s e c 4 x t a n 2 x d x = s e c 4 x t a n x − 4 ∫ s e c 4 x s e c 2 x − s e c 4 x d x = s e c 4 x t a n x − 4 ∫ s e c 6 x d x + 4 ∫ s e c 4 x d x 5 ∫ s e c 6 x d x = s e c 4 x t a n x + 4 ∫ s e c 4 d x ∫ s e c 4 d x = ∫ sec 2 x ( 1 + tan 2 x ) d x = ∫ s e c 2 x d x + ∫ s e c 2 x t a n 2 x d x = tan x + ∫ s e c 2 x t a n 2 x d x ∫ s e c 2 x t a n 2 x d x ( u = t a n x , d u = s e c 2 x d x ) = ∫ u 2 d u = 3 1 u 3 = 3 1 tan 3 x 5 ∫ s e c 6 x d x = s e c 4 x t a n x + 4 ∫ s e c 4 d x = s e c 4 x t a n x + 4 ( t a n x + 3 1 tan 3 x ) ∴ ∫ s e c 6 x d x = 5 1 s e c 4 x t a n x + 5 4 t a n x + 1 5 4 t a n 3 x + C = 5 1 ( 1 + 2 t a n 2 x + t a n 4 x ) t a n x + 5 4 t a n x + 1 5 4 t a n 3 x + C = 5 1 t a n 5 x + 5 5 t a n x + 1 5 1 0 t a n 3 x + C = 5 1 t a n 5 x + 3 2 t a n 3 x + t a n x + C
∫ sec 6 x d x = ∫ ( sec 2 x ) 2 s e c 2 x d x = ∫ ( 1 + tan 2 x ) 2 s e c 2 x d x ( u = t a n x , d u = s e c 2 x d x ) = ∫ ( 1 + u 2 ) 2 d u = ∫ u 4 + 2 u 2 + 1 d u = 1 5 u 5 + 2 3 u 3 + u = 1 5 t a n 5 x + 2 3 t a n 3 x + t a n x + C
\begin{aligned}
&\int \sec^6x dx =\int (\sec^2x)^2 sec^2 x dx=\int (1+\tan^2x)^2 sec^2 x dx\\
&(u=tanx, du=sec^2x dx)\\
&=\int (1+u^2)^2 du = \int u^4+2u^2+1du=\frac{1}{5}u^5+\frac{2}{3}u^3+u\\
&=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\
\end{aligned}
∫ sec 6 x d x = ∫ ( sec 2 x ) 2 s e c 2 x d x = ∫ ( 1 + tan 2 x ) 2 s e c 2 x d x ( u = t a n x , d u = s e c 2 x d x ) = ∫ ( 1 + u 2 ) 2 d u = ∫ u 4 + 2 u 2 + 1 d u = 5 1 u 5 + 3 2 u 3 + u = 5 1 t a n 5 x + 3 2 t a n 3 x + t a n x + C
52. ∫ 1 / ( 5 x − 2 ) 4 d x \int 1/(5x - 2)^4 dx ∫ 1 / ( 5 x − 2 ) 4 d x
∫ 1 ( 5 x − 2 ) 4 d x ( u = 5 x − 2 , d u = 5 d x ) = 1 5 ∫ u − 4 d u = − 1 15 u − 3 = − 1 15 ( 5 x − 2 ) − 3 + C
\begin{aligned}
&\int \frac{1}{(5x - 2)^4}dx (u=5x-2, du=5dx)\\
&=\frac{1}{5}\int u^{-4}du=-\frac{1}{15}u^{-3}=-\frac{1}{15}(5x-2)^{-3}+C
\end{aligned}
∫ ( 5 x − 2 ) 4 1 d x ( u = 5 x − 2 , d u = 5 d x ) = 5 1 ∫ u − 4 d u = − 1 5 1 u − 3 = − 1 5 1 ( 5 x − 2 ) − 3 + C
53. ∫ l n ( 1 + x 2 ) d x \int ln (1+x^2) dx ∫ l n ( 1 + x 2 ) d x
∫ l n ( 1 + x 2 ) d x = ∫ 1 ∗ l n ( 1 + x 2 ) d x = ( l n ∣ 1 + x 2 ∣ ) ( x ) − ∫ 2 x 2 1 + x 2 d x = x l n ∣ 1 + x 2 ∣ − 2 ∫ x 2 + 1 − 1 1 + x 2 d x = x l n ∣ 1 + x 2 ∣ − 2 ∫ 1 − 1 1 + x 2 d x = x l n ∣ 1 + x 2 ∣ − 2 x + 2 ∫ 1 1 + x 2 d x = x l n ∣ 1 + x 2 ∣ − 2 x + 2 arctan x + C
\begin{aligned}
&\int ln (1+x^2)dx =\int 1*ln (1+x^2)dx \\
&=(ln|1+x^2|)(x)-\int \frac{2x^2}{1+x^2} dx\\
&=xln|1+x^2|-2\int \frac{x^2+1-1}{1+x^2} dx\\
&=xln|1+x^2|-2\int 1-\frac{1}{1+x^2} dx\\
&=xln|1+x^2|-2x+2\int\frac{1}{1+x^2} dx\\
&=xln|1+x^2|-2x+2 \arctan{x}+C\\
\end{aligned}
∫ l n ( 1 + x 2 ) d x = ∫ 1 ∗ l n ( 1 + x 2 ) d x = ( l n ∣ 1 + x 2 ∣ ) ( x ) − ∫ 1 + x 2 2 x 2 d x = x l n ∣ 1 + x 2 ∣ − 2 ∫ 1 + x 2 x 2 + 1 − 1 d x = x l n ∣ 1 + x 2 ∣ − 2 ∫ 1 − 1 + x 2 1 d x = x l n ∣ 1 + x 2 ∣ − 2 x + 2 ∫ 1 + x 2 1 d x = x l n ∣ 1 + x 2 ∣ − 2 x + 2 arctan x + C
54. ∫ 1 x 4 + x d x \int \frac{1}{x^4+x} dx ∫ x 4 + x 1 d x
∫ 1 x 4 + x d x = ∫ 1 x ( x 3 + 1 ) d x = ∫ 1 x ( x + 1 ) ( x 2 − x + 1 ) d x = ∫ 1 x + − 1 3 x + 1 + − 2 3 x + 1 3 x 2 − x + 1 d x = l n ∣ x ∣ − 1 3 l n ∣ x + 1 ∣ − 2 3 ∫ x − 1 2 ( x − 1 2 ) 2 + 3 4 d x = l n ∣ x ∣ − 1 3 l n ∣ x + 1 ∣ − 1 3 l n ∣ x 2 − x + 1 ∣ + C = 1 3 l n ∣ x 3 ∣ − 1 3 l n ∣ x + 1 ∣ − 1 3 l n ∣ x 2 − x + 1 ∣ + C = 1 3 l n ∣ x 3 1 ( x + 1 ) 1 ( x 2 − x + 1 ) + C = 1 3 l n ∣ x 3 x 3 + 1 ∣ + C = − 1 3 l n ∣ x 3 + 1 x 3 ∣ + C = − 1 3 l n ∣ 1 + x − 3 ∣ + C
\begin{aligned}
&\int \frac{1}{x^4+x} dx =\int \frac{1}{x(x^3+1)}dx \\
&=\int \frac{1}{x(x+1)(x^2-x+1)}dx \\
&=\int \frac{1}{x}+\frac{-\frac{1}{3}}{x+1}+\frac{-\frac{2}{3}x+\frac{1}{3}}{x^2-x+1}dx \\
&=ln|x|-\frac{1}{3}ln|x+1|-\frac{2}{3} \int \frac{x-\frac{1}{2}}{(x-\frac{1}{2})^2+\frac{3}{4}}dx\\
&=ln|x|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\
&=\frac{1}{3}ln|x^3|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\
&=\frac{1}{3}ln|x^3\frac{1}{(x+1)}\frac{1}{(x^2-x+1)}+C\\
&=\frac{1}{3}ln|\frac{x^3}{x^3+1}|+C=-\frac{1}{3}ln|\frac{x^3+1}{x^3}|+C\\
&=-\frac{1}{3}ln|1+x^{-3}|+C
\end{aligned}
∫ x 4 + x 1 d x = ∫ x ( x 3 + 1 ) 1 d x = ∫ x ( x + 1 ) ( x 2 − x + 1 ) 1 d x = ∫ x 1 + x + 1 − 3 1 + x 2 − x + 1 − 3 2 x + 3 1 d x = l n ∣ x ∣ − 3 1 l n ∣ x + 1 ∣ − 3 2 ∫ ( x − 2 1 ) 2 + 4 3 x − 2 1 d x = l n ∣ x ∣ − 3 1 l n ∣ x + 1 ∣ − 3 1 l n ∣ x 2 − x + 1 ∣ + C = 3 1 l n ∣ x 3 ∣ − 3 1 l n ∣ x + 1 ∣ − 3 1 l n ∣ x 2 − x + 1 ∣ + C = 3 1 l n ∣ x 3 ( x + 1 ) 1 ( x 2 − x + 1 ) 1 + C = 3 1 l n ∣ x 3 + 1 x 3 ∣ + C = − 3 1 l n ∣ x 3 x 3 + 1 ∣ + C = − 3 1 l n ∣ 1 + x − 3 ∣ + C ∫ 1 x 4 + x d x = ∫ 1 x 4 ( 1 + x − 3 ) d x = ∫ x − 4 1 + x − 3 d x ( u = 1 + x − 3 , d u = − 3 x − 4 d x ) = − 1 3 ∫ 1 u d x = − 1 3 l n ∣ u ∣ = − 1 3 l n ∣ 1 + x − 3 ∣ + C
\begin{aligned}
&\int \frac{1}{x^4+x} dx =\int \frac{1}{x^4(1+x^{-3})}dx \\
&=\int \frac{x^{-4}}{1+x^{-3}}dx (u=1+x^{-3}, du=-3x^{-4}dx)\\
&=-\frac{1}{3}\int \frac{1}{u}dx=-\frac{1}{3}ln|u|=-\frac{1}{3}ln|1+x^{-3}|+C
\end{aligned}
∫ x 4 + x 1 d x = ∫ x 4 ( 1 + x − 3 ) 1 d x = ∫ 1 + x − 3 x − 4 d x ( u = 1 + x − 3 , d u = − 3 x − 4 d x ) = − 3 1 ∫ u 1 d x = − 3 1 l n ∣ u ∣ = − 3 1 l n ∣ 1 + x − 3 ∣ + C
55. ∫ 1 − t a n x 1 + t a n x d x \int \frac{1-tan x }{1+tan x} dx ∫ 1 + t a n x 1 − t a n x d x
∫ 1 − t a n x 1 + t a n x d x = ∫ c o s x − s i n x c o s x + s i n x d x = ∫ c o s 2 x − s i n 2 x ( c o s x + s i n x ) 2 d x A l t . = l n ∣ c o s x + s i n x ∣ + C = ∫ c o s ( 2 x ) 1 + s i n ( 2 x ) d x ( u = 1 + s i n ( 2 x ) , d u = 2 c o s ( 2 x ) d x ) = 1 2 ∫ 1 u d u = 1 2 l n ∣ 1 + s i n ( 2 x ) ∣ + C
\begin{aligned}
&\int \frac{1-tan x }{1+tan x} dx \\
&=\int \frac{cos x-sin x }{cos x+sin x} dx =\int \frac{cos^2 x-sin^2 x }{(cos x+sin x)^2} dx \\
Alt. &= ln|cosx+sinx|+C\\
&=\int \frac{cos (2x) }{1+sin(2x)} dx (u=1+sin(2x), du = 2cos(2x)dx)\\
&=\frac{1}{2}\int \frac{1}{u}du\\
&=\frac{1}{2}ln|1+sin(2x)|+C
\end{aligned}
A l t . ∫ 1 + t a n x 1 − t a n x d x = ∫ c o s x + s i n x c o s x − s i n x d x = ∫ ( c o s x + s i n x ) 2 c o s 2 x − s i n 2 x d x = l n ∣ c o s x + s i n x ∣ + C = ∫ 1 + s i n ( 2 x ) c o s ( 2 x ) d x ( u = 1 + s i n ( 2 x ) , d u = 2 c o s ( 2 x ) d x ) = 2 1 ∫ u 1 d u = 2 1 l n ∣ 1 + s i n ( 2 x ) ∣ + C
1 2 l n ∣ 1 + s i n ( 2 x ) ∣ = l n ∣ 1 + 2 s i n x c o s x ∣ = l n ∣ c o s 2 x + s i n 2 x + 2 s i n x c o s x ∣ = l n ∣ c o s x + s i n x ∣
\frac{1}{2}ln|1+sin(2x)|=ln|\sqrt{1+2sinxcosx}|\\
=ln|\sqrt{cos^2x+sin^2x+2sinxcosx}|\\
=ln|cos x + sin x|
2 1 l n ∣ 1 + s i n ( 2 x ) ∣ = l n ∣ 1 + 2 s i n x c o s x ∣ = l n ∣ c o s 2 x + s i n 2 x + 2 s i n x c o s x ∣ = l n ∣ c o s x + s i n x ∣
56. ∫ x ⋅ s e c ( x ) ⋅ t a n ( x ) d x \int x·sec(x)·tan(x) dx ∫ x ⋅ s e c ( x ) ⋅ t a n ( x ) d x
∫ x sec x tan x d x d x ( D s e c → s e c x t a n x ) = x s e c x − ∫ s e c x d x = x s e c x − l n ∣ s e c x + t a n x ∣ + C
\begin{aligned}
&\int x \sec{x} \tan{x} dx dx (D sec \rightarrow sec x tan x) \\
&=x sec x - \int sec x dx\\
&=x sec x - ln|sec x+tan x | + C\\
\end{aligned}
∫ x sec x tan x d x d x ( D s e c → s e c x t a n x ) = x s e c x − ∫ s e c x d x = x s e c x − l n ∣ s e c x + t a n x ∣ + C D ( x sec x − l n ∣ sec x + tan x ∣ ) = sec x + x ( sec x tan x ) − sec x tan x + sec 2 x sec x + tan x = sec x + x sec x tan x − sec x = x sec x tan x
D(x \sec x - ln|\sec x+\tan x | ) \\
= \sec x+x(\sec x \tan x)-\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x } \\
= \sec x + x \sec x \tan x - \sec x = x \sec x \tan x
D ( x sec x − l n ∣ sec x + tan x ∣ ) = sec x + x ( sec x tan x ) − sec x + tan x sec x tan x + sec 2 x = sec x + x sec x tan x − sec x = x sec x tan x
57. ∫ a r c s e c ( x ) d x \int arcsec(x) dx ∫ a r c s e c ( x ) d x
∫ a r c s e c ( x ) d x ( y = a r c s e c ( x ) , x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) = ∫ y s e c ( y ) t a n ( y ) d y = y s e c ( y ) − ∫ s e c ( y ) d y = y s e c ( y ) − l n ∣ s e c ( y ) + t a n ( y ) ∣ ( R . T a n g l e = y , a = 1 , h = x , o = s q r t ( x 2 − 1 ) ) = x a r c s e c ( x ) − l n ∣ x + t a n ( a r c s e c ( x ) ) ∣ + C = x a r c s e c ( x ) − l n ∣ x + x 2 − 1 ∣ + C
\begin{aligned}
&\int arcsec(x) dx \\
&( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\
&=\int y sec(y) tan (y) dy = y sec(y) - \int sec(y) dy \\
&=y sec(y) - ln | sec(y)+tan(y) | \\
&(R.T angle=y, a=1, h=x, o=sqrt(x^2-1) )\\
&=x arcsec(x)-ln |x+tan(arcsec(x)) |+C \\
&=x arcsec(x)-ln |x+\sqrt{x^2-1} |+C \\
\end{aligned}
∫ a r c s e c ( x ) d x ( y = a r c s e c ( x ) , x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) = ∫ y s e c ( y ) t a n ( y ) d y = y s e c ( y ) − ∫ s e c ( y ) d y = y s e c ( y ) − l n ∣ s e c ( y ) + t a n ( y ) ∣ ( R . T a n g l e = y , a = 1 , h = x , o = s q r t ( x 2 − 1 ) ) = x a r c s e c ( x ) − l n ∣ x + t a n ( a r c s e c ( x ) ) ∣ + C = x a r c s e c ( x ) − l n ∣ x + x 2 − 1 ∣ + C
D ( arcsin x ) ( y = a r c s e c ( x ) , x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) = d y d x = 1 s e c ( y ) t a n ( y ) = c o s 2 y csc y = ( 1 x ) 2 x x 2 − 1 = 1 x x 2 − 1
D(\arcsin{x}) \\
( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\
=\frac{dy}{dx} = \frac{1}{sec(y)tan(y)}=cos^2y \csc y\\
=(\frac{1}{x})^2 \frac{x}{\sqrt{x^2-1}}=\frac{1}{x\sqrt{x^2-1}}
D ( arcsin x ) ( y = a r c s e c ( x ) , x = s e c ( y ) , d x = s e c ( y ) t a n ( y ) d y ) = d x d y = s e c ( y ) t a n ( y ) 1 = c o s 2 y csc y = ( x 1 ) 2 x 2 − 1 x = x x 2 − 1 1
∫ a r c s e c ( x ) d x = a r c s e c ( x ) ( x ) − ∫ 1 x x 2 − 1 x d x = x a r c s e c ( x ) − ∫ 1 x 2 − 1 d x
\int arcsec(x) dx = arcsec(x)(x)-\int \frac{1}{x \sqrt{x^2-1}} x dx \\
=x arcsec(x)-\int \frac{1}{\sqrt{x^2-1}}dx
∫ a r c s e c ( x ) d x = a r c s e c ( x ) ( x ) − ∫ x x 2 − 1 1 x d x = x a r c s e c ( x ) − ∫ x 2 − 1 1 d x
∫ 1 x 2 − 1 d x ( x = s e c θ , d x = s e c θ t a n θ d θ ) = ∫ 1 t a n θ s e c θ t a n θ d θ = ∫ s e c θ d θ = l n ∣ s e c θ + t a n θ ∣
\int \frac{1}{\sqrt{x^2-1}}dx (x=sec\theta, dx=sec\theta tan\theta d \theta)\\
=\int \frac{1}{tan \theta} sec\theta tan\theta d \theta\\
=\int sec \theta d \theta =ln|sec \theta+tan \theta|
∫ x 2 − 1 1 d x ( x = s e c θ , d x = s e c θ t a n θ d θ ) = ∫ t a n θ 1 s e c θ t a n θ d θ = ∫ s e c θ d θ = l n ∣ s e c θ + t a n θ ∣
∫ a r c s e c ( x ) d x = x a r c s e c ( x ) − l n ∣ s e c θ + t a n θ ∣ ( R . T a n g l e = θ , a = 1 , h = x , o = s q r t ( x 2 − 1 ) ) ∫ a r c s e c ( x ) d x = x a r c s e c ( x ) − l n ∣ x + x 2 − 1 ∣ + C
\int arcsec(x) dx =x arcsec(x)-ln | sec \theta+tan\theta| \\
(R.T angle=\theta, a=1, h=x, o=sqrt(x^2-1))\\
\int arcsec(x) dx =x arcsec(x)-ln | x+\sqrt{x^2-1}| +C
∫ a r c s e c ( x ) d x = x a r c s e c ( x ) − l n ∣ s e c θ + t a n θ ∣ ( R . T a n g l e = θ , a = 1 , h = x , o = s q r t ( x 2 − 1 ) ) ∫ a r c s e c ( x ) d x = x a r c s e c ( x ) − l n ∣ x + x 2 − 1 ∣ + C
58. ∫ ( 1 − c o s ( x ) ) / ( 1 + c o s ( x ) ) d x \int (1 - cos(x))/(1 + cos(x)) dx ∫ ( 1 − c o s ( x ) ) / ( 1 + c o s ( x ) ) d x
∫ 1 − c o s ( x ) 1 + c o s ( x ) d x = ∫ 1 − 2 c o s ( x ) + c o s 2 x 1 − c o s 2 x d x = ∫ 1 − 2 c o s ( x ) + c o s 2 x s i n 2 x d x = ∫ 1 s i n 2 x d x − 2 ∫ c o s x s i n 2 x d x + ∫ c o s 2 x s i n 2 x d x
\begin{aligned}
&\int \frac{1 - cos(x)}{1 + cos(x)}dx \\
&=\int \frac{1-2cos(x)+cos^2x}{1-cos^2x} dx =\int \frac{1-2cos(x)+cos^2x}{sin^2x} dx \\
&=\int \frac{1}{sin^2x}dx-2\int \frac{cos x}{sin^2 x}dx+\int \frac{cos^2x}{sin^2x} dx
\end{aligned}
∫ 1 + c o s ( x ) 1 − c o s ( x ) d x = ∫ 1 − c o s 2 x 1 − 2 c o s ( x ) + c o s 2 x d x = ∫ s i n 2 x 1 − 2 c o s ( x ) + c o s 2 x d x = ∫ s i n 2 x 1 d x − 2 ∫ s i n 2 x c o s x d x + ∫ s i n 2 x c o s 2 x d x
∫ 1 s i n 2 x d x = ∫ c o s 2 x + s i n 2 x s i n 2 x = ∫ c o s 2 x s i n 2 x d x + x = ∫ c o t 2 x d x + x = ∫ c s c 2 x − 1 d x + x = ∫ c s c 2 x d x = − c o t x ∫ c o s x s i n 2 x d x ( u = s i n x , d u = c o s x d x ) = ∫ d u u 2 = − 1 u = − c s c x ∫ c o s 2 x s i n 2 x d x = ∫ c o t 2 x d x = ∫ c s c 2 x − 1 d x = − c o t x − x
\int \frac{1}{sin^2x} dx=\int \frac{cos^2x + sin^2x}{sin^2x}\\
=\int \frac{cos^2x}{sin^2x} dx +x = \int cot^2x dx +x\\
=\int csc^2x-1 dx+x =\int csc^2x dx = -cot x \\
\int \frac{cos x}{sin^2x} dx (u=sin x, du=cos x dx)\\
=\int \frac{du}{u^2} = -\frac{1}{u}=-csc x\\
\int \frac{cos^2x}{sin^2x} dx =\int cot^2x dx =\int csc^2x-1dx\\
=-cot x - x
∫ s i n 2 x 1 d x = ∫ s i n 2 x c o s 2 x + s i n 2 x = ∫ s i n 2 x c o s 2 x d x + x = ∫ c o t 2 x d x + x = ∫ c s c 2 x − 1 d x + x = ∫ c s c 2 x d x = − c o t x ∫ s i n 2 x c o s x d x ( u = s i n x , d u = c o s x d x ) = ∫ u 2 d u = − u 1 = − c s c x ∫ s i n 2 x c o s 2 x d x = ∫ c o t 2 x d x = ∫ c s c 2 x − 1 d x = − c o t x − x
= ∫ 1 s i n 2 x d x − 2 ∫ c o s x s i n 2 x d x + ∫ c o s 2 x s i n 2 x d x = − c o t x + 2 c s c x − c o t x − x = 2 c s c x − 2 c o t x − x + C = 2 ( c s c x − c o t x ) − x + C = 2 ( 1 − c o s x s i n x ) − x + C
=\int \frac{1}{sin^2x}dx-2\int \frac{cos x}{sin^2 x}dx+\int \frac{cos^2x}{sin^2x} dx \\
= -cot x +2 csc x -cot x - x\\
=2 csc x -2cot x -x +C =2 (csc x -cot x) -x +C\\
=2(\frac{1-cos x}{sin x})-x+C\\
= ∫ s i n 2 x 1 d x − 2 ∫ s i n 2 x c o s x d x + ∫ s i n 2 x c o s 2 x d x = − c o t x + 2 c s c x − c o t x − x = 2 c s c x − 2 c o t x − x + C = 2 ( c s c x − c o t x ) − x + C = 2 ( s i n x 1 − c o s x ) − x + C
= 2 ( 1 − c o s x 2 s i n x 2 ) − x + C = 2 ( s i n 2 x 2 2 s i n x 2 c o s x 2 / 2 ) − x + C = 2 t a n ( x 2 ) − x + C
=2(\frac{\frac{1-cosx}{2}}{\frac{sinx}{2}})-x+C\\
=2(\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}/2} )-x+C\\
=2tan(\frac{x}{2})-x+C
= 2 ( 2 s i n x 2 1 − c o s x ) − x + C = 2 ( 2 s i n 2 x c o s 2 x / 2 s i n 2 2 x ) − x + C = 2 t a n ( 2 x ) − x + C
Alt.∫ 1 − c o s ( x ) 1 + c o s ( x ) d x ( s i n 2 ( x 2 ) = 1 − c o s x 2 , c o s 2 ( x 2 ) = 1 + c o s x 2 ) = ∫ s i n 2 ( x 2 ) c o s 2 ( x 2 ) d x = ∫ 1 − c o s 2 ( x 2 ) c o s 2 ( x 2 ) d x = ∫ s e c 2 x 2 d x − x = 2 t a n ( x 2 ) − x + C
\begin{aligned}
&\int \frac{1 - cos(x)}{1 + cos(x)}dx \\
&( sin^2(\frac{x}{2})=\frac{1-cosx}{2} , cos^2(\frac{x}{2})=\frac{1+cosx}{2} )\\
&=\int \frac{sin^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx =\int \frac{1-cos^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx \\
&=\int sec^2{\frac{x}{2}} dx-x \\
&=2tan(\frac{x}{2})-x+C
\end{aligned}
∫ 1 + c o s ( x ) 1 − c o s ( x ) d x ( s i n 2 ( 2 x ) = 2 1 − c o s x , c o s 2 ( 2 x ) = 2 1 + c o s x ) = ∫ c o s 2 ( 2 x ) s i n 2 ( 2 x ) d x = ∫ c o s 2 ( 2 x ) 1 − c o s 2 ( 2 x ) d x = ∫ s e c 2 2 x d x − x = 2 t a n ( 2 x ) − x + C
59. ∫ ( x 2 ) s q r t ( x + 4 ) d x \int (x^2)sqrt(x + 4)dx ∫ ( x 2 ) s q r t ( x + 4 ) d x
∫ x 2 x + 4 d x ( u = x + 4 , d u = 1 2 x + 4 d x ) = ∫ ( u 2 − 4 ) 2 u 2 u d u = ∫ 2 u 2 ( u 4 − 8 u 2 + 16 ) d u = ∫ 2 u 6 − 16 u 4 + 32 u 2 d u = 2 7 u 7 − 16 5 u 5 + 32 3 u 3 + C = 2 7 ( x + 4 ) 7 2 − 16 5 ( x + 4 ) 5 2 + 32 3 ( x + 4 ) 3 2 + C
\begin{aligned}
&\int x^2\sqrt{x+4}dx \\
&(u=\sqrt{x+4}, du=\frac{1}{2\sqrt{x+4}}dx)\\
&=\int(u^2-4)^2u 2u du=\int 2u^2(u^4-8u^2+16)du\\
&=\int 2u^6-16u^4+32u^2 du\\
&=\frac{2}{7}u^7-\frac{16}{5}u^5+\frac{32}{3}u^3+C\\
&=\frac{2}{7}(x+4)^\frac{7}{2}-\frac{16}{5}(x+4)^\frac{5}{2}+\frac{32}{3}(x+4)^\frac{3}{2}+C\\
\end{aligned}
∫ x 2 x + 4 d x ( u = x + 4 , d u = 2 x + 4 1 d x ) = ∫ ( u 2 − 4 ) 2 u 2 u d u = ∫ 2 u 2 ( u 4 − 8 u 2 + 1 6 ) d u = ∫ 2 u 6 − 1 6 u 4 + 3 2 u 2 d u = 7 2 u 7 − 5 1 6 u 5 + 3 3 2 u 3 + C = 7 2 ( x + 4 ) 2 7 − 5 1 6 ( x + 4 ) 2 5 + 3 3 2 ( x + 4 ) 2 3 + C
60. ∫ − 1 1 s q r t ( 4 − x 2 ) d x \int_{-1}^1 sqrt(4 - x^2) dx ∫ − 1 1 s q r t ( 4 − x 2 ) d x
∫ − 1 1 4 − x 2 d x = 2 ∫ 0 1 4 − x 2 d x ( x = 2 s i n θ , d x = 2 c o s θ d θ ) = 2 ∫ 0 π / 6 2 c o s θ 2 c o s θ d θ = 8 ∫ c o s 2 θ d θ = 4 ∫ 1 + c o s 2 θ d θ = 4 θ + 2 s i n 2 θ = 4 ( π / 6 ) + 2 ( 3 / 2 ) = 2 π 3 + 3
\begin{aligned}
&\int_{-1}^1 \sqrt{4 - x^2} dx = 2\int_0^1 \sqrt{4 - x^2} dx \\
&( x=2sin\theta, dx=2cos\theta d\theta)\\
&=2\int_0^{\pi/6} 2cos \theta2cos\theta d\theta = 8\int cos^2 \theta d\theta\\
&=4\int 1+cos2\theta d\theta = 4\theta+2sin2\theta \\
&=4(\pi/6)+2(\sqrt{3}/2)\\
&=\frac{2\pi}{3}+\sqrt{3}
\end{aligned}
∫ − 1 1 4 − x 2 d x = 2 ∫ 0 1 4 − x 2 d x ( x = 2 s i n θ , d x = 2 c o s θ d θ ) = 2 ∫ 0 π / 6 2 c o s θ 2 c o s θ d θ = 8 ∫ c o s 2 θ d θ = 4 ∫ 1 + c o s 2 θ d θ = 4 θ + 2 s i n 2 θ = 4 ( π / 6 ) + 2 ( 3 / 2 ) = 3 2 π + 3
Author: crazyj7@gmail.com
