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

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

Answers (1)

Answer:

\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}

 

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