Mathematics

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


ANSWER

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


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
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