IIT-JEE 1981Question 1
Evaluate $\displaystyle\int\left(e^{\log x}+\sin x\right)\cos x\,dx$.
Indefinite Integration
Answer$x\sin x+\cos x-\dfrac14\cos2x+C$
Solution
Concept. $e^{\log x}=x$ (for $x>0$), so the integral is $\int x\cos x\,dx+\int\sin x\cos x\,dx$. Formulas used. (1) $\int x\cos x\,dx=x\sin x+\cos x$ (by parts)(2) $\sin x\cos x=\frac12\sin2x$, $\int\frac12\sin2x\,dx=-\frac14\cos2x$Step 1 - Simplify. The integrand is $x\cos x+\sin x\cos x$. Step 2 - Use (1), (2). $$x\sin x+\cos x-\frac14\cos2x+C.$$ Answer: $x\sin x+\cos x-\dfrac14\cos2x+C$