Common Integration Trick

Calculus Level 3

Evaluate: π π n = 1 9 sin ( n x ) ln ( x 2 + e ) d x \Large \int_{-\pi}^{\pi} \large \frac{\sum\limits_{n=1}^{9} {\sin(nx)}}{\ln(x^2+e)}dx


The answer is 0.

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2 solutions

Marta Reece
Dec 25, 2017

Sine is an odd function. Sum of several odd functions is odd. Dividing it by an even function leaves the function odd.

Integral of an odd function over an interval symmetrical with respect to the origin is zero.

Chew-Seong Cheong
Dec 25, 2017

I = π π n = 1 9 sin ( n x ) ln ( x 2 + e ) d x = n = 1 9 π π sin ( n x ) ln ( x 2 + e ) d x Note that sin ( n x ) is odd and ln ( x 2 + e ) is even, = n = 1 9 0 therefore the integrand sin ( n x ) ln ( x 2 + e ) is odd. = 0 \begin{aligned} I & = \int_{-\pi}^\pi \frac {\sum_{n=1}^9 \sin (nx)}{\ln (x^2+e)} dx \\ & = \sum_{n=1}^9 \color{#3D99F6} \int_{-\pi}^\pi \frac {\sin (nx)}{\ln (x^2+e)} dx & \small \color{#3D99F6} \text{Note that }\sin (nx) \text{ is odd and }\ln(x^2+e) \text{ is even, } \\ & = \sum_{n=1}^9 \color{#3D99F6} 0 & \small \color{#3D99F6} \text{therefore the integrand }\frac {\sin (nx)}{\ln (x^2+e)} \text{ is odd.} \\ & = \boxed{0} \end{aligned}

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