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Q.
$\int \frac{\sin ^{2} x \cdot \sec ^{2} x+2 \tan x \cdot \sin ^{-1} x \cdot \sqrt{1-x^{2}}}{\sqrt{1-x^{2}}\left(1+\tan ^{2} x\right)} d x=$
Integrals
Solution:
$\int \frac{\sin ^{2} x \cdot \sec ^{2} x+2 \tan x \cdot \sin ^{-1} x \cdot \sqrt{1-x^{2}}}{\sqrt{1-x^{2}}\left(1+\tan ^{2} x\right)} d x$
$=\int\left(\frac{\sin ^{2} x}{\sqrt{1-x^{2}}}+\frac{2 \tan x \cdot \sin ^{-1} x}{\sec ^{2} x}\right) d x$
$=\int \frac{\sin ^{2} x}{\sqrt{1-x^{2}}} d x+\int 2 \sin x \cos x \sin ^{-1} x d x$
$=\sin ^{2} x \sin ^{-1} x-\int 2 \sin x \cos x \sin ^{-1} x d x+\int 2 \sin x \cos x \sin ^{-1} x d x+ C$
$=\sin ^{2} x \sin ^{-1} x+C$