Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
योग
Advertisements
उत्तर
\[\text{ Let I } = \int\frac{1}{\sin^2 x + \sin 2x}dx\]
\[ = \int\frac{1}{\sin^2 x + 2 \sin x \cdot \cos x}dx\]
Dividing numerator and denominator by cos2x, we get
\[I = \int\frac{\frac{1}{\cos^2 x}}{\tan^2 x + 2 \tan x}dx\]
\[ = \int\frac{\sec^2 x}{\tan^2 x + 2 \tan x} dx\]
\[\text{ Putting tan x = t}\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\]
\[ \therefore I = \int\frac{1}{t^2 + 2t}dt\]
\[ = \int\frac{1}{t^2 + 2t + 1 - 1}dt\]
\[ = \int\frac{1}{\left( t + 1 \right)^2 - 1^2}dt\]
\[ = \frac{1}{2} \text{ ln} \left| \frac{t + 1 - 1}{t + 1 + 1} \right| + C\]
\[ = \frac{1}{2} \text{ ln } \left| \frac{t}{t + 2} \right| + C \]
\[ = \frac{1}{2} \text{ ln} \left| \frac{\tan x}{\tan x + 2} \right| + C ............\left[ \because t = \tan x \right]\]
\[ = \int\frac{\sec^2 x}{\tan^2 x + 2 \tan x} dx\]
\[\text{ Putting tan x = t}\]
\[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\]
\[ \therefore I = \int\frac{1}{t^2 + 2t}dt\]
\[ = \int\frac{1}{t^2 + 2t + 1 - 1}dt\]
\[ = \int\frac{1}{\left( t + 1 \right)^2 - 1^2}dt\]
\[ = \frac{1}{2} \text{ ln} \left| \frac{t + 1 - 1}{t + 1 + 1} \right| + C\]
\[ = \frac{1}{2} \text{ ln } \left| \frac{t}{t + 2} \right| + C \]
\[ = \frac{1}{2} \text{ ln} \left| \frac{\tan x}{\tan x + 2} \right| + C ............\left[ \because t = \tan x \right]\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int x e^x \text{ dx }\]
\[\int x \cos^2 x\ dx\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\]
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\sqrt{x^2 - a^2} \text{ dx}\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
