#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 42 maths textbook solution

$\pi$

Hint:

We use $\int_{-a}^{a} f(x) d x=0$ function.

Given:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$

Explanation:

\begin{aligned} &\int_{-a}^{a} f(x) d x=0\\\\ &f(x) \text { odd function }\\\\ &f(-x)=-f(x)\\\\ &(-x)^{3}=-x^{3} \ldots(i) \end{aligned}

\begin{aligned} &x \cos x \\\\ &(-x) \cos (-x)=-x \cos x \ldots(i i) \\\\ &\tan ^{5} x \\\\ &\tan ^{5}(-x)=-\tan ^{5} x \ldots(i i i) \end{aligned}

\begin{aligned} &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{3} d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan ^{5} x d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d x \\\\ &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d x \end{aligned}

\begin{aligned} &=[x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\\\ &=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right) \\\\ &=\frac{2 \pi}{2} \\\\ &=\pi \end{aligned}