Mathematics

# $\displaystyle \int \dfrac{\sin x \cos x }{\sqrt{1 - \sin^4 x}}dx$  is equal to

$\dfrac{1}{2} \sin^{-1}(\sin^2 x) + C$

##### SOLUTION
$\int \dfrac{\sin x \cos x}{\sqrt{1-\sin^{4} x} } dx$
Put $\sin x = t \Rightarrow \cos x dx = dt$
$\Rightarrow \int \dfrac{t dt}{\sqrt{1-t^{4}}}$
Again, put $t^{2}=y \Rightarrow 2t dt = dy \Rightarrow t dt = \dfrac{dy}{2}$
$\Rightarrow \int \dfrac{1}{\sqrt{1-y^{2}}} \dfrac{dy}{2}$
$= \dfrac{1}{2} \int \dfrac{1}{\sqrt{1-y^{2}}} dy$ Again, put $y=\sin \theta$
$\Rightarrow dy= \cos \theta d \theta$
$= \dfrac{1}{2} \int \dfrac{1 (\cos \theta)}{\sqrt{1-\sin^{2} \theta}} d \theta$
$= \dfrac {1}{2} \int \dfrac{\cos \theta}{\cos \theta} d \theta = \dfrac{1}{2} \int d \theta =\dfrac{1}{2} \theta$
Putting back the substituted values, we get
$=\dfrac{1}{2} \sin^{-1} (y)$
$=\dfrac{1}{2} \sin^{-1} (t^{2})$
$=\dfrac{1}{2} \sin^{-1} (\sin^{2} x)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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