$\int \sin x(\cos x+7)^{2} d x=-\frac{(\cos x+7)^{3}}{3}+C$ .

Name: $\int \sin x(\cos x+7)^{2} d x=-\frac{(\cos x+7)^{3}}{3}+C$ .
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Quizplus · Accepted Answer

The integral is solved using substitution. Let $ u = \cos x + 7 $, then $ du = -\sin x \, dx $. The integral becomes $-\int u^2 \, du$, which evaluates to $-\frac{u^3}{3} + C$. Substituting back $ u = \cos x + 7 $ gives the result.

∫sin⁡x(cos⁡x+7)2dx=−(cos⁡x+7)33+C\int \sin x(\cos x+7)^{2} d x=-\frac{(\cos x+7)^{3}}{3}+C∫sinx(cosx+7)2dx=−3(cosx+7)3​+C

$\int \sin x(\cos x+7)^{2} d x=-\frac{(\cos x+7)^{3}}{3}+C$