Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
Advertisements
उत्तर
\[Let, I = \int_0^\frac{\pi}{4} e^x \sin x d x ..............(1)\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} e^x \cos x dx\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} e^x \sin x dx\]
\[ \Rightarrow I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - I ..............\left[\text{Using (1)} \right] \]
\[ \Rightarrow 2I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} \]
\[ = - \frac{1}{\sqrt{2}} e^\frac{\pi}{4} + 1 + \frac{1}{\sqrt{2}} e^\frac{\pi}{4} - 0\]
\[ = 1\]
\[\text{Hence }I = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
Γ(4)
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Find: `int logx/(1 + log x)^2 dx`
