Mathematics

# $\int \frac{1}{sin^{2}xcos^{2}x}dx$

##### SOLUTION
$\int \frac{1}{sin^{2}x.cos^{2}x} dx \Rightarrow sin^{2}x.cos^{2}x$  $sin\,2x = 2\,sin\,x.cos\,x$
$= (sinx.cosx)^{2}$  $\frac{1}{2} sin\,2x = sin\,x.cos\,x$
$= (\frac{1}{2} sin\,2x)^{2}$
$= \frac{1}{4} sin^{2}\,2x$
$= \int \frac{1}{\frac{1}{4}sin^{2}2x}dx$
$= \int (\frac{1}{4} sin^{2}2x)^{-1}dx$
$= \int 4 \,cosec^{2}\,2x.dx$  $\int \,cosec^{2}\,x=-cotx\,$
$= -4\,cot\,2x \times \frac{1}{2} + c$
$= -2\,cot \,2x + c$

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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 $\int_{0}^{1} x \tan^{-1} x\ dx$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle \frac{3x+4}{(x+1)^2(x-1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$, then $A$=
• A. $\displaystyle \frac{1}{2}$
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• C. $\displaystyle \frac{1}{4}$
• D. $\displaystyle \frac{7}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
$\displaystyle \int_{0}^{\pi/2}\sqrt{1+cosx}dx$

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Q4 Single Correct Medium
If $f (x) =\displaystyle \int_{2x}^{\sin x } \cos (t^3) dt ,$ then $f'(x)$ is equal to :
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• B. $\cos ( \cos^3 x) \cos x -2 \cos ( 8x^3)$
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$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$