Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} d x . Then, \]
\[I = \int_0^2 \frac{1}{\sqrt{- x^2 + 2x - 1 + 1 + 3}} d x\]
\[ \Rightarrow I = \int_0^2 \frac{1}{\sqrt{- \left( x - 1 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 1 \right)}{2} \right]_0^2 \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} \left( - \frac{1}{2} \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} \frac{1}{2}\]
\[ \Rightarrow I = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Solve each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
