Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 10

Answers (1)

Answer:

        \frac{1}{3 a^{3}} \tan ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C

Hint:

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

Given:

        \int \frac{x^{2}}{x^{6}+a^{6}}dx

Solution:

Let\: \: \: I=\int \frac{x^{2}}{x^{6}+a^{6}}dx

                =\int \frac{x^{2}}{(x^{3})^{2}+(a^{3})^{2}}dx

Put \: \: x^{3}=t\Rightarrow 3x^{2}dx=dt\Rightarrow x^{2}dx=\frac{dt}{3}

Then\: \: I=\int \frac{1}{t^{2}+(a^{3})^{2}}dt

                \begin{aligned} &=\frac{1}{3} \int \frac{1}{t^{2}+\left(a^{3}\right)^{2}} d t \\ &=\frac{1}{3} \times \frac{1}{a^{3}} \tan ^{-1}\left(\frac{t}{a^{3}}\right)+C \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \\ &=\frac{1}{3 a^{3}} \tan ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C \quad\left[\because t=x^{3}\right] \end{aligned}

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads