Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
उत्तर
\[\int_0^\frac{\pi}{3} \frac{\cos x}{3 + 4\sin x} d x\]
\[Let, \sin x = t \Rightarrow \cos x dx = dt\]
\[\text{When, }\sin x \to 0 ; t \to 0\]
\[\text{And }\sin x \to \frac{\pi}{3} ; t \to \frac{\sqrt{3}}{2}\]
\[ = \int_0^\frac{\sqrt{3}}{2} \frac{dt}{3 + 4t}\]
\[ = \frac{1}{4}\log \left[ 3 + 4t \right]_0^\frac{\sqrt{3}}{2} \]
\[ = \frac{1}{4}\log\left[ \log\left( 3 + 2\sqrt{3} \right) - \log\left( 3 + 0 \right) \right]\]
\[ = \frac{1}{4}\log\left[ \log\left( 2\sqrt{3} + 3 \right) - \log\left( 3 \right) \right]\]
\[ = \frac{1}{4}\left( \log\frac{2\sqrt{3} + 3}{3} \right)\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
