Mathematics

# Integrate:$\int \dfrac{\cos 4x}{\sin^2x}dx$

##### SOLUTION
$I=\displaystyle\int\dfrac{\cos 4x}{\sin^{2}x}dx$

$=\displaystyle\int \dfrac{\cos^{2}2x-\sin^{2}2x}{\sin^{2}x}dx$

$=\displaystyle\int\dfrac{(\cos^{2}x-\sin^{2}x)^2-4\sin^{2}x\cos^{2}x}{\sin^{2}x}dx$

$=\displaystyle\int\dfrac{\cos^{4}x+\sin^{4}x-6\sin^{2}x\cos^{2}x}{\sin^{2}x}dx$

$=\displaystyle\int\dfrac{\sin^{4}x+\cos^{2}x(1-7\sin^{2}x)dx}{\sin^{2}x}$

$[\because \sin^{2}x+\cos^{2}x=1]$

$\displaystyle\int \sin^{2}xdx+\int\dfrac{\cos^{2}x(1-7\sin^{2}x)dx}{\sin^{2}x}$

$=I_{1}+I_{2}$ ............. $(1)$

$I_{1}=\displaystyle\int\sin^{2}x dx=\int \dfrac{(1-\cos 2x)}{2}dx$

$=\dfrac{x}{2}-\dfrac{\sin 2x}{4}$ ....... $(2)$

$I_{2}=\displaystyle\int\dfrac{\cos^{2}x(1-7\sin^{2}x)}{\sin^{2}x}dx$

$=\displaystyle\int\cot^{2}dxd-7\int \cos^{2}x dx$

$=\displaystyle\int (\csc^{2}x-1)dx-\dfrac{7}{2}\int (1+\cos 2x)dx$

$=-\cot x-x\dfrac{-7}{2}x-\dfrac{7}{4}\sin 2x$

$=-\cot x-\dfrac{7}{4}\sin 2x-\dfrac{9}{2}x$ ........ $(3)$

$I=I_{1}+I_{2}$

$\Rightarrow I=-\cot x-2\sin 2x-4x+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium
Evaluate $\displaystyle \int \frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}} dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve $\displaystyle\int { \dfrac { { e }^{ 6\log { x } }-{ e }^{ 5\log { x } } }{ { e }^{ 4\log { x } }-{ e }^{ 3\log { x } } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int\frac{\sin 2x}{a^{2}+b^{2}\sin^{2}x}dx=$
• A. $\dfrac{1}{a^{2}-b^{2}}\log|a^{2}+b^{2}\sin^{2}x|+c$
• B. $\dfrac{1}{a^{2}}\log|a^{2}+b^{2}\sin^{2}x|+c$
• C. $\displaystyle \frac{1}{a^{2}+b^{2}}\log|a^{2}+b^{2}\sin^{2}x|+c$
• D. $\displaystyle \frac{1}{b^{2}}\log|a^{2}+b^{2}\sin^{2}x|+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium

Evaluate the following definite integral:

$\displaystyle \int_{0}^{1} \ x+x^2 dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$