Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 + \cos 2x \right) dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ \Rightarrow I = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{4} + 0 - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
`Γ (9/2)`
`int x^3/(x + 1)` is equal to ______.
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
