$\int_{-a}^{a}\left(x^{3}+x \cos ^{2} 2 x+ an ^{3} x+3 ight) d x=$\

Question

$\int_{-a}^{a}\left(x^{3}+x \cos ^{2} 2 x+	an ^{3} x+3ight) d x=$\

Professor PC Thomas Classes · Answer

(d): We have, $\int_{-a}^{a}\left(x^{3}+x \cos ^{2} 2 x+	an ^{3} x+3ight) d x$\ $=\int_{-a}^{a} x^{3} d x+\int_{-a}^{a} x \cos ^{2} 2 x d x+\int_{-a}^{a} 	an ^{3} x d x+\int_{-a}^{a} 3 d x$\ $\begin{array}{ccc}\underset{\left(	ext { let } I_{1}ight)}{\downarrow} & \underset{\left(	ext { let } I_{2}ight)}{\downarrow} & \underset{\left(	ext { let } I_{3}ight)}{\downarrow}\end{array}$\ Now, $I_{1}=\int_{-a}^{a} x^{3} d x=0 \quad\left(\because I_{1}ight.$ is an odd function $)$\ $I_{2}=\int_{-a}^{a} x \cos ^{2} 2 x d x=0 \quad\left(\because I_{2}ight.$ is an odd function $)$\ $I_{3}=\int_{-a}^{a} 	an ^{3} x d x=0 \quad\left(\because I_{3}ight.$ is an odd function $)$\ $	herefore \quad \int_{-a}^{a}\left(x^{3}+x \cos ^{2} 2 x+	an ^{3} x+3ight) d x=\int_{-a}^{a} 3 d x=3(x)_{-a}^{a}$\ $=3 	imes 2 a=6 a$\