Q111 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

If $\int e^{\sin ^2 x}\left(\sin x \cos x+\cos ^3 x \sin x\right) d x$ $=e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

  • A
    $\frac{1}{2} \sin ^2 x$
  • B
    $\frac{1}{2} \cos ^2 x$
  • C
    $-\frac{1}{2} \cos 2 x$
  • Correct answer
    D
    $-\frac{1}{2} \sin 2 x$

Solution

$I=\int e^{\sin ^2 x}\left(\sin x \cos x+\cos ^3 x \sin x\right) d x$ $=\int e^{\sin ^2 x}\left(1+\cos ^2 x\right) \sin x \cos x d x$ $=\int e^{\sin ^2 x}\left(2-\sin ^2 x\right) \sin x \cos x d x$ Let $\sin ^2 x=t \Rightarrow \sin x \cos x d x=\frac{d t}{2}$ $\therefore I=\frac{1}{2} \int e^t(2-t) d t=e^t-\frac{1}{2}\left[t \cdot e^t-e^t\right]+C$ $=e^t\left[1-\frac{t}{2}+\frac{1}{2}\right]+C=e^{\sin ^2 x}\left[1+\frac{1}{2}-\frac{\sin ^2 x}{2}\right]+C$ $=e^{\sin ^2 x}[1+f(x)]+C$(given) Therefore, $f(x)=\frac{1}{2}-\frac{\sin ^2 x}{2} \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sin 2 x$.

Hence, the correct answer is option: 'D'.

Relevant Questions