$\int \theta^{2} \cos (a \theta) d \theta=\frac{\theta^{2}}{a} \sin (a \theta)+\frac{2 \theta}{a^{2}} \cos (a \theta)+C$ , where a is a constant.

Question

Quizplus · Accepted Answer

The given integral is incorrect. The integration by parts method should be applied twice to solve $\int \theta^{2} \cos (a \theta) d \theta$, and the correct result involves terms with both sine and cosine functions, but the coefficients and terms differ from those given.

∫θ2cos⁡(aθ)dθ=θ2asin⁡(aθ)+2θa2cos⁡(aθ)+C\int \theta^{2} \cos (a \theta) d \theta=\frac{\theta^{2}}{a} \sin (a \theta)+\frac{2 \theta}{a^{2}} \cos (a \theta)+C∫θ2cos(aθ)dθ=aθ2​sin(aθ)+a22θ​cos(aθ)+C

$\int \theta^{2} \cos (a \theta) d \theta=\frac{\theta^{2}}{a} \sin (a \theta)+\frac{2 \theta}{a^{2}} \cos (a \theta)+C$