JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q17

MathematicsSingle correct+4 / -1Question 17 of 75

The value of $\displaystyle\int_0^{20\pi}\left(\sin^4x+\cos^4x\right)dx$ is equal to :

\dfrac{15\pi}2

25\pi

15\pi

\dfrac{25\pi}2

Answer and worked solution

Official answer: C

Simplify. $\sin^4x+\cos^4x=1-\frac12\sin^22x=\frac34+\frac14\cos4x$ .

Integrate. Over $[0,20\pi]$ the cosine term integrates to 0: $\frac34\cdot20\pi=15\pi$ .

(C)

JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 17