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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 40

Answers (1)

Answer:

0

Hint:

Given:

\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x

Explanation:

\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1-x-x^{2}}\right) d x \\\\ &=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x \end{aligned}

=\int_{0}^{1} \tan ^{-1} x-\tan ^{-1}(1-x) d x                .........Eq(i)

Using properties,

\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\\\ &I=\int_{0}^{1} \tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x)) d x \end{aligned}

=\int_{0}^{1} \tan ^{-1}(1-x)-\tan ^{-1} x \; d x            ..........Eq(ii)

Adding   Eq(i)  and  Eq(ii)

\begin{aligned} &2 I=0 \\\\ &I=0 \end{aligned}

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