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Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 42.

Answers (1)

Answer   : 9 \sqrt{3}-1

Hint   : use indefinite formula and the limit to solve this integral

Given  \int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta

Solution  : \int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta

Put (5-4\cos \theta )=t\Rightarrow -4(-\sin \theta )d\theta =dt  

\rightarrow \sin \theta d\theta = \frac{dt}{4}

when \theta =0 then t=1 and when \theta =\pi then t=9


\int_{0}^{\pi}5(5-4 \cos \theta)^{\frac{1}{4}}\sin \theta d \theta

=\int_{1}^{9} 5 t^{\frac{1}{4}} \frac{d t}{4}=\frac{5}{4} \int_{1}^{9} t^{\frac{1}{4}} d t

\begin{aligned} &=\frac{5}{4}\left[\frac{t^{\frac{1}{4}+1}}{\frac{1}{4}+1}\right]_{1}^{9} \\ &=\frac{5}{4}\left[\frac{t^{\frac{5}{4}}}{\frac{5}{4}}\right]_{1}^{9} \\ &=\frac{5}{4} \times \frac{4}{5}\left[9^{\frac{5}{4}}-1^{\frac{5}{4}}\right] \\ &=3^{2 x_{4}^{5}}-1=3^{\frac{5}{2}}-1 \\ &=9 \sqrt{3}-1 \end{aligned}


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