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  • Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 40 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 40 Maths Textbook Solution.

Answers (1)

Answer:

           \frac{2}{\sqrt{7}}\left [ \tan ^{-1}\frac{5}{\sqrt{7}}-\tan ^{-1}\frac{1}{\sqrt{7}} \right ]

Hint: Use indefinite formula and then put limits to solve this integral

Given:

\int_{0}^{1}\frac{1}{2x^{2}+x+1}dx

Solution:

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

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

=\frac{1}{2} \int_{0}^{1} \frac{1}{x^{2}+2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x

=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x

=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\frac{1}{16}+\frac{1}{2}} d x

=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1-8}{16}\right)} d x

=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{-7}{16}\right)} d x

=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d x

Put

x+\frac{1}{4}=u\Rightarrow dx=du

When x=0 then

u=\frac{1}{4}

when x=1 then

u=1+\frac{1}{4}=\frac{4+1}{4}=\frac{5}{4}

Then

\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{1}{2 x^{2}+x+1} d x=\frac{1}{2} \int_{\frac{1}{4}}^{\frac{5}{4}} \frac{1}{u^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d u \\ &=\frac{1}{2}\left[\frac{\frac{1}{\sqrt{7}}}{4} \tan ^{-1} \frac{u}{\frac{\sqrt{7}}{4}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \\ &=\frac{1}{2} \times \frac{4}{\sqrt{7}}\left[\tan ^{-1} \frac{4 u}{\sqrt{7}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \end{aligned}

\begin{aligned} &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{4 \times \frac{5}{4}}{\sqrt{7}}-\tan ^{-1} \frac{4 \times \frac{1}{4}}{\sqrt{7}}\right] \\ &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1} \frac{1}{\sqrt{7}}\right] \end{aligned}

 

Posted by

infoexpert21

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