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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 51 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 51 Maths Textbook Solution.

Answers (1)

Answer:0

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

Given:

\int_{0}^{2\pi }e^\frac{x}{2}\sin \left ( \frac{x}{2}+\frac{\pi }{4} \right )dx

Sol:

\int_{0}^{2\pi }e^\frac{x}{2}\sin \left ( \frac{x}{2}+\frac{\pi }{4} \right )dx

=\int_{0}^{2 \pi} e^{\frac{x}{2}}\left[\sin \frac{\pi}{4} \cos \frac{x}{2}+\cos \frac{\pi}{4} \sin \frac{x}{2}\right] d x

[\because \sin (A+B)=\sin A \cos B+\cos A \sin B]

=\int_{0}^{2 \pi} e^{\frac{x}{2}}\left[\frac{1}{\sqrt{2}} \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}\right] d x

=\frac{1}{\sqrt{2}} \int_{0}^{2 \pi}\left[\cos \frac{x}{2} e^{\frac{x}{2}}+\sin \frac{x}{2} e^{\frac{x}{2}}\right] d x

=\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} \cos \frac{x}{2} e^{\frac{x}{2}} d x+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} \sin \frac{x}{2} e^{\frac{x}{2}} d x

=\frac{1}{\sqrt{2}}\left\{\int_{0}^{2 \pi} e^{\frac{x}{2}} \sin \frac{x}{2} d x\right\}+\frac{1}{\sqrt{2}}\left\{\int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x\right\}

=\frac{1}{\sqrt{2}}\left\{\left[\sin \frac{x}{2} \frac{e^{\frac{x}{2}}}{\frac{1}{2}}\right]_{0}^{2 \pi}-\int_{0}^{2 \pi} \cos \frac{x}{2} \cdot \frac{1}{2} \frac{e^{\frac{x}{2}}}{\frac{1}{2}} d x\right\}+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x

[using integration by parts]

=\frac{1}{\sqrt{2}}\left[2 \sin \frac{x}{2} e^{\frac{x}{2}}\right]_{0}^{2 \pi}-\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x

=\frac{1}{\sqrt{2}} \times 2\left[\sin \frac{2 \pi}{2} \cdot e^{\frac{2 \pi}{2}}-\sin \frac{0}{2} \cdot e^{\frac{0}{2}}\right]

=\sqrt{2}\left[\sin \pi \cdot e^{\pi}-\sin 0 \cdot e^{0}\right]

=\sqrt{2}\left [ 0-0 \right ]

                                                                        [\because \sin \pi=\sin 0=0]

=0

Posted by

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