Question

$\int e^{\sin \theta }[\log \sin \theta +\cos e{c}^2\theta ]\cos \theta d\theta$ is equal to

• A. $\int e^{\sin \theta }[\log \sin \theta +\cos e{c}^2\theta ]+c$
• B. $e^{\sin \theta }[\log \sin \theta +\cos ec\theta ]+c$
• C. $e^{\sin \theta }[\log \sin \theta -\cos ec\theta ]+c$
• D. $e^{\sin \theta }[\log \sin \theta -\cos e{c}^2\theta ]+c$
• E. $e^{\sin \theta }[\log \sin \theta +{\cos }^2\theta ]+c$

Answer 1

Answer: (C)

Let $I=\int e^{\sin \theta }\log \sin \theta \cos \theta d\theta$
$+\int e^{\sin \theta }\cos e{c}^2\theta \cos \theta d\theta$ Put $\sin \theta =t\Rightarrow \cos \theta d\theta =dt$
$\therefore$ $I=\int e^t\log tdt+\int \frac{e^t}{t}dt+\frac{e^t{t}^{-1}}{-1}-\int \frac{e^t{t}^{-1}}{-1}dt$
$=e^t\left(\log t-\frac{1}{t}\right)+c$
$=e^{\sin \theta }(\log \sin \theta -\cos ec\theta )+c$