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  • Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 1 maths textbook solution

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 1 maths textbook solution

Answers (1)

Answer:

        \frac{1}{2}log\left | \frac{1+tan\, x}{1-tan\, x} \right |+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

        \int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x

Solution:

Let\: \: \int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x

Put\: \: tan\: x=t\Rightarrow sec^{2}x\: dx=dt

Then \: \: \: I=\int \frac{1}{1-t^{2}} d t=\int \frac{1}{1^{2}-t^{2}} d t

                    \begin{array}{ll} =\frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+C \quad & {\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right]} \\ \\ =\frac{1}{2} \log \left|\frac{1+\tan x}{1-\tan x}\right|+C & {[\because t=\tan x]} \end{array}

Posted by

Gurleen Kaur

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