Get Answers to all your Questions

header-bg qa

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

Answers (1)


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


\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}\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]


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support