Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{1 + \sin x} d x . Then, \]
\[I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1 - \sin x}{1 - \sin^2 x} dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1 - \sin x}{\cos^2 x} dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \sec^2 x - \sec x \tan x \right) dx\]
\[ \Rightarrow I = \left[ \tan x - \sec x \right]_{- \frac{\pi}{4}}^\frac{\pi}{4} \]
\[ \Rightarrow I = \left( 1 - \sqrt{2} \right) - \left( - 1 - \sqrt{2} \right)\]
\[ \Rightarrow I = 2\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
