41. $\int \sinh{x}dx$

$\begin{aligned} &\int \sinh {x} dx \\ &=\int \frac {e^x-e^{-x}}{2} dx =\frac{1}{2}(\int e^xdx-\int e^{-x}dx)\\ &=\frac{e^x+e^{-x}}{2}=\cosh {x}+C \end{aligned}$

42. $\int \sinh^2{x}dx$

$\begin{aligned} &\int \sinh^2 {x} dx \\ &=\int (\frac {e^x-e^{-x}}{2})^2 dx =\frac{1}{4}(\int e^{2x}dx+\int e^{-2x}dx-\int 2 dx)\\ &=\frac{e^{2x}-e^{-2x}}{8}-\frac{1}{2}x=\frac{1}{4}\sinh {2x}-\frac{1}{2}x+C \end{aligned}$

43. $\int \sinh^3{x}dx$

$\begin{aligned} &\int \sinh^3{x} dx \\ &=\int \sinh{x}\sinh^2{x} dx =\int \sinh{x}(\cosh^2{x}-1) dx\\ &=\int sinhx cosh^2x dx -\int sinh x dx (u=cosh x, du = sinh x dx)\\ &=\int u^2 du-cosh x=\frac{1}{3}cosh^3x-coshx+C \end{aligned}$

44. $\int \frac{1}{\sqrt{1+x^2}}dx$

$\begin{aligned} &\int \frac{1}{\sqrt{1+x^2}}dx (x=tan(y), dx=sec^2y dy)\\ &=\int \frac{1}{sec{y}} sec^2ydy=\int \sec{y} dy\\ &=ln|\sec{y}+\tan{y}|+C\\ &(R.T. x=tan y, a=1, o=x, h=\sqrt{1+x^2})\\ &=ln|\sqrt{1+x^2}+x|+C\\ \end{aligned}$
Alternative

let $\sinh^{-1}{x}=y$, $sinh{y}=x$
$=ln|\sqrt{1+sinh^2y}+sinh{y}|\\ =ln|\sqrt{1+(\frac{e^y-e^{-y}}{2})^2}+\frac{e^y-e^{-y}}{2}|\\ =ln|\sqrt{\frac{e^{2y}+e^{-2y}-2+4}{4}}+\frac{e^y-e^{-y}}{2}|\\ =\ln| \frac{e^y+e^{-y}}{2} +\frac{e^y-e^{-y}}{2}| = \ln |e^y|=y\\ =\sinh^{-1}{x}$

45. $\int ln(x + sqrt(x^2 + 1) ) dx$

$\begin{aligned} &\int ln(x + \sqrt{x^2 + 1} )dx = \int sinh^{-1}x dx\\ &(x=sinh \theta, dx=cosh\theta d\theta) \\ &=\int ln (sinh \theta +\sqrt{sinh^2 \theta+1})cosh \theta d \theta\\ &=\int ln (sinh \theta + cosh \theta)cosh\theta d \theta \\ &=\int \theta cosh \theta d \theta \\ &=\theta sinh \theta - cosh\theta+C\\ &=x \sinh^{-1}{x}-\sqrt{sinh^2\theta+1}+C\\ &=x \sinh^{-1}{x}-\sqrt{x^2+1}+C \end{aligned}$

46. $\int \tanh{x} dx$

$\begin{aligned} &\int \tanh{x} dx \\ &=\int \frac{sinh x}{cosh x} dx, (u=cosh x, du = sinh x dx )\\ &=\int \frac{du}{u} = ln |u| +C=ln|cosh x|+C \end{aligned}$

47. $\int sech{x} dx$

(sech{x})’ = -sech(x)tanh(x)
$\begin{aligned} &\int sech{x} dx = \int \frac{1}{cosh{x}} dx = \int \frac{cosh{x}}{cosh^2{x}} dx \\ &=\int \frac{coshx}{sinh^2x+1}dx (u=sinhx, du=coshxdx)\\ &=\int \frac{1}{1+u^2} du\\ &=\arctan {u} = \tan^{-1}({\sinh^{-1}x})+C \end{aligned}$

48. $\int \tanh^{-1}{x} dx$

(y=tanh^-1 x, x=tanh y, dx=sech^2 y dy)
(tanh x)’ = sech^2(x)
$\begin{aligned} &\int \tanh^{-1}{x} dx \\ &=\int y sech^2y dy = y tanh(y)-\int tanh(y) dy\\ &=y tanh(y)-ln|cosh(y)|\\ &=xtanh^{-1}x-ln|cosh(tanh^{-1}x)|+C \\ \end{aligned}$

$y=tanh^{-1}x \\ cosh(tanh^{-1}x) =cosh (y)\\ ... ??$

$\begin{aligned} &\int \tanh^{-1}{x} dx \\ &(tanh^{-1}x \rightarrow D \rightarrow \frac{1}{1-x^2})\\ &=(\tanh^{-1}{x}) (x) - \int \frac{1}{1-x^2}xdx\\ &=x \tanh^{-1}{x} - \int \frac{x}{1-x^2}dx (u=1-x^2, du=-2xdx)\\ &=x \tanh^{-1}{x} +\frac{1}{2} \int \frac{1}{u} du\\ &=x \tanh^{-1}{x} +\frac{1}{2}ln|1-x^2|+C \end{aligned}$

49. $\int \sqrt{\tanh{x}} dx$

$\begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &=\int \sqrt{\frac{sinh x}{cosh x}} dx\\ &(u=\sqrt{coshx}, du =\frac{sinhx}{2\sqrt{cosh x}}dx )\\ &=\int \frac{\sqrt{sinh x}}{u}\frac{2\sqrt{coshx}}{sinhx}du\\ &=2\int \frac{1}{\sqrt{sinhx}}du=2\int \frac{1}{\sqrt{ \sqrt{cosh^2x-1}}}du\\ &=2\int \frac{1}{(u^4-1)^{1/4}}du \end{aligned}$

$\begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &=\int \sqrt{\frac{sinh x}{cosh x}} dx\\ &(u=coshx, du =sinhx dx )\\ &=\int \frac{\sqrt{sinh x}}{\sqrt{u}}\frac{1}{sinhx}du\\ &=\int \frac{1}{\sqrt{u}\sqrt{sinhx}}du=\int \frac{1}{\sqrt{u\sqrt{u^2-1} }}du\\ &=\int\frac{1}{(u^4-u^2)^{1/4}}du \end{aligned}$

$\begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &(u=\sqrt{tanh(x)}, u^2=tanhx, x=arctanh(u^2) )\\ &(dx=\frac{1}{1-u^4}2udu)\\ &=\int u \frac{1}{1-u^4}2udu \\ &=2\int \frac{u^2}{1-u^4}du = 2\int \frac{u^2}{(1-u^2)(1+u^2)}du\\ &=2\int \frac{\frac{1}{2}}{1-u^2}+\frac{-\frac{1}{2}}{1+u^2} du\\ &=\int \frac{1}{1-u^2}-\frac{1}{1+u^2} du\\ &=arctanh(u)-arctan(u)\\ &=arctanh(\sqrt{tanh(x)})-arctan(\sqrt{tanh(x)})+C\\ \end{aligned}$

50. $\int_0^5 [x] dx$

$\begin{aligned} &\int_0^5 [x] dx \\ &x=[0,1) y=0, x=[1,2)y=1, x=[2,3)y=2...\\ &Area=0+1+2+3+4=10 \end{aligned}$

Author: crazyj7@gmail.com

31. $\int \frac{1}{\sqrt{x-x^{3/2}}}dx$

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^{3/2}}}dx\\ &=\int \frac{1}{\sqrt{x}\sqrt{1-x^{1/2}}}dx\\ &=\int \frac{x^{-1/2}}{\sqrt{1-x^{1/2}}}dx\\ &(u=1-x^{1/2}, du=-\frac{1}{2}x^{-1/2} dx) \\ &=-2\int \frac{1}{u^{1/2}} du=-2\int u^{-1/2} du\\ &=-2 (2)u^{1/2}=-4\sqrt{u}\\ &=-4\sqrt{1-\sqrt{x}}+C \end{aligned}$

32. $\int \frac{1}{\sqrt{x-x^2}}dx$

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^2}}dx \\ &=\int \frac{1}{x\sqrt{x^{-1}-1}} dx =\int \frac{x^{-1}}{\sqrt{x^{-1}-1}} dx\\ &(u=x^{-1}-1, du=-x^{-2}dx)\\ &=-\int \frac{x^{-1}}{\sqrt{u}}x^2du \end{aligned}$

$\begin{aligned} &\int \frac{1}{\sqrt{x-x^2}}dx \\ &=\int \frac{1}{\sqrt{x}\sqrt{1-x}} dx \\ &(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx)\\ &=\int \frac{1}{u\sqrt{1-u^2}} 2udu\\ &=2\int \frac{1}{\sqrt{1-u^2}}du \\ \end{aligned}$
Right Triangle : h=1, o=u, a=sqrt(1-u^2) ,sin$\theta$ = u
$=2\int \frac{1}{\cos \theta} \cos \theta d\theta =2\theta = 2 \arcsin{u}+C\\ =2\arcsin{\sqrt{x}}+C$

33. $\int e^{2lnx} dx$

$\begin{aligned} &\int e^{2lnx} dx\\ &=\int e^{lnx}e^{lnx} dx =\int (e^{lnx})^2 dx =\int x^2 dx\\ or&=\int e^{lnx^2} dx =\int x^2 dx\\ &=\frac{1}{3}x^3+C \end{aligned}$

34. $\int lnx/sqrt x dx$

$\begin{aligned} &\int \frac{\ln x}{\sqrt x}dx \\ &(u=\sqrt x , du = \frac{1}{2\sqrt x}dx)\\ &=2\int ln u^2 du=4\int ln u du \\ &=4 (uln |u| -u ) \\ &=4 (\sqrt x ln |\sqrt x| - \sqrt x ) + C\\ &=2\sqrt x ln (x) - 4\sqrt x + C\\ \end{aligned}$

$\int ln x dx = (ln x) x - \int 1/x * x dx = x(lnx)-x$

35. $\int \frac{1}{e^x+e^{-x}} dx$

$\begin{aligned} &\int \frac{1}{e^x+e^{-x}} dx \\ & \text{we know} \; cosh x=\frac{e^x+e^{-x}}{2}\\ &=\frac{1}{2}\int \frac{1}{cosh x} dx \\ &=\int \frac{e^x}{e^{2x}+1} dx (u=e^x, du=e^xdx)\\ &=\int \frac{du}{u^2+1} =\arctan {u}\\ &=\arctan{e^x}+C \end{aligned}$

36. $\int log_2 x dx$

$\begin{aligned} &\int log_2 x dx =\int \frac{ln x}{ln 2} dx\\ &=\frac{1}{ln 2}\int ln x dx\\ &=\frac{1}{ln 2}(x ln x - x)+C\\ &=x log_2x - \frac{x}{ln 2} + C\\ \end{aligned}$

37. $\int x^3*sin(2x) dx$

$\begin{aligned} &\int x^3\sin{2x} dx \\ &=x^3(-\frac{1}{2}cos2x)-3x^2(-\frac{1}{4}sin2x)+6x(\frac{1}{8}cos2x)-6(\frac{1}{16}sin2x)\\ &=cos 2x (-\frac{1}{2}x^3+\frac{3}{4}x)+sin2x(\frac{3}{4}x^2-\frac{3}{8})+C \end{aligned}$

38. $\int x^2[1+x^3]^{1/3} dx$

$\begin{aligned} &\int x^2 \sqrt[3]{1+x^3} dx \\ &(u=1+x^3, du=3x^2dx) \\ &=\frac{1}{3}\int \sqrt[3]u du = \frac{1}{3}\int u^{\frac{1}{3}} du \\ &=\frac{1}{3} \frac{3}{4}u^{1+\frac{1}{3}}=\frac{1}{4}u\sqrt[3]{u}\\ &=\frac{1}{4}(1+x^3)\sqrt[3]{1+x^3}+C \end{aligned}$

39. $\int 1/(x^2 + 4)^2 dx$

$\begin{aligned} &\int \frac{1}{(x^2 + 4)^2} dx \\ &(x=2tany, dx=2sec^2ydy, y=\arctan{\frac{x}{2}}) \\ &=\int \frac{2sec^2y}{(4(tan^2y+1))^2}dy=\int \frac{sec^2y}{8sec^4y}dy\\ &=\frac{1}{8}\int cos^2y dy =\frac{1}{16}\int 1+\cos{2y}dy\\ &=\frac{y}{16}+\frac{1}{32}\sin{2y}=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}sin y cos y\\ &(right triangle angle=y, h=sqrt(x^2+4) a=2, o=x)\\ &=\frac{1}{16}arctan{\frac{x}{2}}+\frac{1}{16}\frac{x}{\sqrt{x^2+4}} \frac{2}{\sqrt{x^2+4}}\\ &=\frac{1}{16}arctan{\frac{x}{2}}+\frac{x}{8(x^2+4)}+C \end{aligned}$

40. $\int_1^2 sqrt(x^2-1) dx$

$\begin{aligned} &\int_1^2 \sqrt{x^2-1} dx , (x=sec(y), dx=sec(y) tan(y) dy)\\ &\int \sqrt{\tan^2y} \sec y \tan y dy\\ &=\int \sec y \tan^2 y dy (sec y tan y -I-> sec y) \\ &=\tan y \sec y -\int \sec^3 y dy\\ \end{aligned}$

$\int \sec^3 x dx = \int \sec x \sec^2 x dx (sec^2x-I->tan x)\\ =\sec x \tan x - \int \sec x \tan x \tan x dx\\ =\sec x \tan x - \int \sec x (\sec^2 x -1 ) dx \\ =\sec x \tan x - \int \sec^3 x dx +\int \sec x dx \\ =\sec x \tan x + ln |sec x + tan x|-\int \sec^3x dx\\ = \frac{1}{2}(\sec x \tan x + ln |sec x + tan x|)$

x=sec(y), y=arcsec x, RT. angle=y, h=x, a=1, o=sqrt(x^2-1)
$=\tan y \sec y -\int \sec^3 y dy\\ =\tan y \sec y -\frac{1}{2}(\sec y \tan y + ln |sec y + tan y|)\\ =\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|+C\\ \int_1^2 \sqrt{x^2-1} dx=\left[\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}ln| \sqrt{x^2-1}+x|\right]_1^2\\ =\sqrt{3}-\frac{1}{2}ln(\sqrt{3}+2) \\$

Author: crazyj7@gmail.com

21. $\int \sin^3{x} \cos^2{x}dx$

$\begin{aligned} &\int \sin^3{x} \cos^2{x}dx\\ &=\int \sin{x}\sin^2{x}\cos^{2}x dx\\ &=\int \sin{x}(1-\cos^2{x})\cos^{2}x dx\\ &(u=cosx, du=-sinx dx) \\ &=-\int (1-u^2)u^{2} du = \int u^4-u^2dx\\ &=\frac{1}{5}cos^5x-\frac{1}{3}cos^3x+C\\ \end{aligned}$

22. $\int \frac{1}{x^2\sqrt{x^2+1}} dx$

$\begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{tan^2\theta sec\theta} sec^2\theta d\theta (x=\tan{\theta}, dx=sec^2\theta d\theta) \\ &=\int sec\theta cot^2\theta d\theta =\int \frac{cos^2\theta}{cos\theta sin^2\theta} d\theta \\ &=\int \frac{cos \theta}{sin^2 \theta} d\theta (t=sin\theta, dt=cos\theta d\theta)\\ &=\int \frac{dt}{t^2} = -\frac{1}{t}=-\frac{1}{sin\theta}=-csc\theta+C\\ &=-\csc({\arctan{x}}) + C \\ &=-\frac{\sqrt{x^2+1}}{x} + C \\ \end{aligned}$
Alternative

$\begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{x^2 x \sqrt{1+x^{-2}}} dx\\ &=\int \frac{x^{-3}}{\sqrt{1+x^{-2}}} dx (u=1+x^{-2}, du=-2x^{-3}dx)\\ &=\int -\frac{1}{2}u^{-1/2}du = -\frac{1}{2}2u^{1/2}=-\sqrt{1+\frac{1}{x^2}}+C\\ \end{aligned}$

23. $\int \sin{x}\sec{x}\tan{x} dx$

$\begin{aligned} &\int \sin{x}\sec{x}\tan{x} dx = \int \tan^2x dx =\int sec^2x -1dx\\ &=\int \frac{1-cos^2x}{cos^2x} dx = \int sec^2x dx-x\\ &=\tan{x}-x+C\\ \end{aligned}$

24. $\int sec^3(x)dx$

$\begin{aligned} &\int sec^3(x)dx=\int sec(x)sec^2(x) dx\\ &= \sec x \tan x - \int \sec x \tan x \tan x dx\\ &= \sec x \tan x - \int \sec x \tan^2 x dx\\ &= \sec x \tan x - \int \sec x (sec^2x-1) dx\\ &= \sec x \tan x + \int sec x dx - \int sec^3x dx\\ &= \sec x \tan x + \ln | secx+tanx| - \int sec^3x dx \\ \end{aligned}$

$\begin{aligned} &2\int sec^3(x)dx=\sec x \tan x +\ln | secx+tanx| \\ &\therefore \int sec^3(x)dx=\frac{1}{2} (\sec x \tan x +\ln | secx+tanx|)+C \\ \end{aligned}$

25. $\int 1/(x*sqrt(9x^2-1)) dx$

$\begin{aligned} &\int \frac{1}{x\sqrt{9x^2-1}} dx \\ &(3x=\sec{y}, 3dx=\sec y \tan y dy)\\ &=\int \frac{3}{\sec{y} \tan{y} } \frac{\sec y \tan y }{3} dy \\ &= y =\sec^{-1} 3x +C \\ \end{aligned}$

26. $\int cos(sqrt(x)) dx$

$\begin{aligned} &\int \cos ({\sqrt{x}}) dx \\ &(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx )\\ &= \int \cos {u} 2 \sqrt{x} du\\ &= 2\int u\cos {u} du = 2( u sin u - (-cos u) ) \\ &= 2u sin u +2 cos u +C\\ &=2\sqrt{x}\sin{\sqrt{x}} + 2 \cos{\sqrt{x}} + C \\ \end{aligned}$

27. $\int \cosec{x}dx$

$\begin{aligned} &\int \cosec{x} dx \\ &=\int \frac{\cosec{x}(cosec{x}+cot{x}) }{ (cosec{x}+cot{x}) } dx \\ &( u = cosec{x}+cot{x} , du = (-\cosec{x}\cot{x}-\cosec^2{x}) dx )\\ &=-\int \frac{1}{u} du \\ &=-\ln|{\cosec{x}+\cot{x}}| + C \\ \end{aligned}$

28. $\int sqrt(x^2+4x+13) dx$

$\begin{aligned} &\int \sqrt{x^2+4x+13} dx \\ &=\int \sqrt{(x+2)^2+3^2} dx \\ & (u=\frac{x+2}{3} , 3 du = dx) \\ &=3\int \sqrt{3^2u^2+3^2} du \\ &=9\int \sqrt{u^2+1} du \\ &(u=tan {y}, du=sec^2ydy)\\ &=9\int \sec{y} \sec^2{y} dy = 9\int sec^3y dy \\ &=9\left[sec(y)tan(y)-\int sec(y)tan(y)tan(u) dy\right] \\ &=9sec(y)tan(y)-9\int sec(y)(sec^2y-1) dy \\ &=9sec(y)tan(y)-9\int sec^3(y)dy+9\int sec y dy \\ &18\int sec^3(y)dy = 9sec(y)tan(y)+9\ln|sec(y)+tan(y)|\\ &9\int sec^3(y)dy = \frac{9}{2}(sec(y)tan(y)+\ln|sec(y)+tan(y)|)\\ &= \frac{9}{2}sec(y)tan(y)+\frac{9}{2}\ln|sec(y)+tan(y)|\\ &(tan{y}=\frac{x+2}{3}, angle=y, h=\sqrt{x^2+4x+13} , adj=3, opposite=x+2)\\ &= \frac{9}{2}\frac{\sqrt{x^2+4x+13}}{3}\frac{x+2}{3}+\frac{9}{2}\ln{|\frac{\sqrt{x^2+4x+13}}{3}+\frac{x+2}{3}|} \\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|\frac{x+2+\sqrt{x^2+4x+13}}{3}|}+C\\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|x+2+\sqrt{x^2+4x+13}|}+C_2\\ \end{aligned}\\$

29. $\int e^{2x}*cosx dx$

$\begin{aligned} &\int e^{2x} \cos{x} dx \\ &=e^{2x} sin{x}-(2e^{2x})(-cos{x})+\int (4e^{2x})(-cos{x}) dx \\ &=e^{2x}sin{x}+2e^{2x}cos{x}-4\int e^{2x}cos{x}dx\\ &5\int e^{2x}\cos{x} dx= e^{2x}sin{x}+2e^{2x}cos{x} \\ &\int e^{2x} \cos{x} dx =\frac{1}{5}e^{2x}sin{x}+ \frac{2}{5}e^{2x}cos{x}+C\\ \end{aligned}$

30. $\int_3^5 (x-3)^9 dx$

$\begin{aligned} &\int_3^5 (x-3)^9 dx (u=x-3)\\ &=\int_0^2 u^9 du =\bigg[ \frac{u^{10}}{10} \bigg ]_0^2 \\ &=102.4 \end{aligned}$

Author: crazyj7@gmail.com