Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_0^{2\pi} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} - \frac{2}{2} e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]
\[\text{Again, integrating second term by parts}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left\{ \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} \frac{2}{2} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x \right\}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - I\]
\[ \Rightarrow 2I = \frac{2}{\sqrt{2}} e^\pi + \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} e^\pi - \frac{2}{\sqrt{2}} = 0\]
\[ \Rightarrow I = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.