Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{6} \cos x \cos 2x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{6} \cos x \left( \cos^2 x - \sin^2 x \right) dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{6} \left( 2 \cos^3 x - \cos x \right) dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{6} \left( 2 \cos x \left( 1 - \sin^2 x \right) - \cos x \right) dx\]
\[ \Rightarrow I = \left[ 2\left( \sin x - \frac{\sin^3 x}{3} \right) - \sin x \right]_0^\frac{\pi}{6} \]
\[ \Rightarrow I = \left[ 2\left( \frac{1}{2} - \frac{1}{24} \right) - \frac{1}{2} \right] - 0\]
\[ \Rightarrow I = \frac{5}{12}\]
APPEARS IN
संबंधित प्रश्न
Prove that:
Evaluate :
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
