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# Can someone explain - Integral Calculus - JEE Main-17

The value of $\int_{0}^{2\pi}[sin2x(1+cos3x)]dx$, where $[t]$ denotes

greatest integer function, is :

• Option 1)

$\pi$

• Option 2)

$-\pi$

• Option 3)

$-2\pi$

• Option 4)

$2\pi$

Views

$I=$ $\int_{0}^{2\pi}[sin2x(1+cos3x)]dx$.....................(1)

$I=$ $\int_{0}^{2\pi}[sin2(2\pi-x)(1+cos3(2\pi-x))]dx$

$I=$ $\int_{0}^{2\pi}[-sin2x(1+cos3x)]dx$..............................(2)

Add (1) and (2)

$2I=$ $\int_{0}^{2\pi}([sin2x(1+cos3x)]+[-sin2x(cos 3x+1)])dx$

$2I=$ $\int_{0}^{2\pi}-1dx$

$2I=$ $[-x]_{0}^{2\pi}$

$2I=$ $-2\pi$

$I=$ $-\pi$

So, option (2) is correct.

Option 1)

$\pi$

Option 2)

$-\pi$

Option 3)

$-2\pi$

Option 4)

$2\pi$

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