Home › JEE Advanced › Mathematics › Definite Integrals › $\displaystyle\int_0^{\pi/2} \sin^2 x\, dx = ?$
$\displaystyle\int_0^{\pi/2} \sin^2 x\, dx = ?$
A$\pi/4$
B$\pi/2$
C$1$
D$\pi^2/8$
Answer & Solution
Correct answer: A. $\pi/4$
$\sin^2 x = (1 - \cos 2x)/2$. $\int_0^{\pi/2} (1 - \cos 2x)/2\, dx = \tfrac{1}{2}[x - \tfrac{1}{2}\sin 2x]_0^{\pi/2} = \tfrac{1}{2}(\pi/2 - 0) = \pi/4$.
Related questions
$\displaystyle\int_0^{\infty} \dfrac{\ln x}{1 + x^2}\, dx = ?$$\displaystyle\int_0^1 \dfrac{x - 1}{\ln x}\, dx = ?$ (Frullani-type)Given $\displaystyle\int_0^{\pi/2} \dfrac{ in^n x}{ in^n x + \cos^n x}\, dx$, the value isIf $\displaystyle f(x) = \int_0^x t in t\, dt$, then $f''(\pi) = ?$$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$Evaluate $\displaystyle\lim_{n\to\infty} \dfrac{1^k + 2^k + \cdots + n^k}{n^{k+1}}$ for $k$\displaystyle\int_0^\pi \dfrac{x}{1 + in x}\, dx = ?$$\displaystyle\int_0^{\pi/4} \tan^4 x\, dx = ?$