Advertisements
Advertisements
Question
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
Sum
Advertisements
Solution
\[\text{ Let I } = \int\frac{1}{\sin^2 x + \sin 2x}dx\]
\[ = \int\frac{1}{\sin^2 x + 2 \sin x \cdot \cos x}dx\]
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
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
\[\int \cos^2 \frac{x}{2} dx\]
Integrate the following integrals:
\[\int\text { sin x cos 2x sin 3x dx}\]
\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\]
\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int x^3 \cos x^4 dx\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int x \text{ sin 2x dx }\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int\left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int \sec^6 x\ dx\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
