Integral100 [91-100]

91. $\int \frac{x}{1 + x^4} dx$

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

---

92. $\int e^{\sqrt x} dx$

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

---

93. $\int \frac{1}{csc(x)^3} dx$

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

---

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

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

---

95. $\int \sqrt{1 + sin(2x)} dx$

$\begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=sin(2x) , du=2cos(2x)dx\\ &=\int \sqrt {1+u} \frac{1}{2cos(2x)}du =\frac 1 2 \int \sqrt {1+u} \frac {1}{1-2sin^2(x)}du \\ &=\frac 1 2 \int \sqrt u \frac{1}{1-2(u-1)^2}du =\frac 1 2 \int \sqrt u \frac{1}{-2u^2+4u-1}du \\ & not work \end{aligned}$

$\begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=\sqrt{1 + sin(2x)}, du=\frac{2cos(2x)}{2\sqrt{1 + sin(2x)}}dx\\ &u^2-1=sin(2x)\\ &=\int u\frac{u}{cos(2x)} du = \int \frac{u^2}{\sqrt{1-(u^2-1)^2}}du\\ & t=u^2-1, dt=2udu\\ &=\frac{1}{2} \int \frac{u}{\sqrt{1-t^2}}dt=\frac{1}{2}\int \frac{\sqrt{1+t}}{\sqrt{1-t}\sqrt{1+t}}dt \\ &=\frac{1}{2}\int \frac{1}{\sqrt{1-t}}dt=\frac{1}{2}\int (1-t)^{-1/2}dt=\frac 1 2 (2)(1-t)^{1/2}\\ &=\sqrt{1-t}=\sqrt{2-u^2}=\sqrt{2-(1+sin(2x))}\\ &=\sqrt{1-sin(2x)}+C\\ &=\sqrt{cos^2x+sin^2x-2sinxcos}+C=|cosx-sinx|+C \end{aligned}$
Alt.
$1=sin^2x+cos^2x\\ \int \sqrt{1 + sin(2x)} dx =\int \sqrt {sin^2x+cos^2x+2sinxcosx} dx\\ =\int (sinx+cosx)dx=-cosx+sinx+C$

---

96. $\int x^{1/4} dx$

$\begin{aligned} &\int x^{1/4} dx \\ &=\frac 4 5 x^{\frac 5 4}+C \end{aligned}$

---

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

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

---

98. $\int \sqrt{1 + e^x} dx$

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

---

99. $\int \frac{\sqrt{tan(x)}}{sin(2x)}dx$

$\begin{aligned} &\int \frac {\sqrt{tan(x)}}{sin(2x)}dx =\int \frac {\sqrt{tan(x)}}{2sin(x)cos(x)} dx\\ &u=\sqrt {tan(x)}, du=\frac{sec^2x}{2\sqrt{tanx}}dx \\ &=\int \frac {u}{2sinxcosx}\frac{2u}{sec^2x}du \\\\ &=\int \frac{2tan(x)cos(x)}{2sinx} du \\ &=\int du= u =\sqrt{tan(x)}+C \end{aligned}$

---

100. $\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx$

$\begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ & \text{div cos} \\ &=\int \frac{secx}{secx+tanx}dx\\ & u=secx+tanx, du=(secxtanx+sec^2x )dx\\ &=\int \frac{secx}{u} \frac{1}{secx(tanx+secx)}du\\ &=\int \frac{1}{u^2}du = -\frac{1}{u}\\ &=-\frac{1}{secx+tanx}+C=-\frac{cosx}{1+sinx}+C \\ & -\frac{cosx}{1+sinx} ]_0^{\pi/2} =0-(-1)\\ &=1 \end{aligned}$
Alt.
$\begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ &=\int_0^{\pi/2} \frac{1-sin(x)}{1 - sin^2(x)} dx =\int \frac{1-sin(x)}{cos^2(x)} dx\\ &=\int sec^2x-sec(x)tan(x)dx=tan(x)-sec(x)+C\\ &=tan(x)-sec(x) ]_0^{\pi/2} not solve. \\ &=\frac{sin(x)-1}{cos(x)}=-\frac{cos^2(x)}{cos(x)(1+sin(x))}=-\frac{cos(x)}{1+sin(x)} \end{aligned}$

---

101. $\int \frac {sin(x)} x + ln(x)cos(x) dx$

$\begin{aligned} &\int \frac{sin(x)}{x} + ln(x)cos(x) dx \\ &=\int \frac{sinx}{x}dx+\int ln(x)cos(x)dx\\ &=sin(x)ln(x)- \int cos(x)ln(x)dx+\int ln(x)cos(x)dx\\ &=sin(x)ln(x)+C \end{aligned}$

---

---

Author: crazyj7@gmail.com