Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\, I = \int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[ = \int_0^{\pi/2} \frac{x + \sin x}{2 \cos^2 \frac{x}{2}} dx\]
\[ = \int_0^{\pi/2} \left[ \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{\sin x}{2 \cos^2 \frac{x}{2}} \right]dx\]
\[ = \frac{1}{2} \int_0^{\pi/2} x se c^2 \frac{x}{2}dx + \int_0^{\pi/2} \frac{2\sin \frac{x}{2}\cos\frac{x}{2}}{2 \cos^2 \frac{x}{2}}dx\]
\[ = \frac{1}{2} \left[ x \frac{\tan\frac{x}{2}}{\frac{1}{2}} \right]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{\tan\frac{x}{2}}{\frac{1}{2}}dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ x \tan\frac{x}{2} \right]_0^{\pi/2} - \int_0^{\pi/2} \tan\frac{x}{2} dx + \int_0^{\pi/2} \tan\frac{x}{2}dx\]
\[ = \left[ \frac{\pi}{2} \tan\frac{\pi}{4} \right]\]
\[ = \frac{\pi}{2} \times 1\]
\[ = \frac{\pi}{2} \]
APPEARS IN
संबंधित प्रश्न
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Solve each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Find: `int logx/(1 + log x)^2 dx`
