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

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

Answers (1)

best_answer

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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