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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 39

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 39

Answers (1)

Answer: \frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c

Hint: Use substitution method to solve this integral.

Given:   \int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x

Solution:

        \begin{aligned} &\text { Let } I=\int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x \\ &\text { Put } x^{2}+3=t \Rightarrow 2 x d x=d t \\ &\Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}

        \begin{aligned} \Rightarrow I &=\int 2 x \sec ^{3} t \cdot \tan t \frac{d t}{2 x}=\int \sec ^{3}(t) \cdot \tan (t) d t \\ &=\int \sec ^{2}(t) \sec (t) \tan (t) d t \end{aligned}

        \text { Again Put } \sec t=u \Rightarrow \sec t \tan t \; d t=d u \text { then }

        I=\int u^{2} d u=\left[\frac{u^{2+1}}{2+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

           \begin{array}{ll} =\frac{u^{3}}{3}+c=\frac{\sec ^{3} t}{3}+c & {[\because \sec t=u]} \end{array}

           =\frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c \quad\left[\because t=x^{2}+3\right]

 

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