Advertisements
Advertisements
प्रश्न
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
बेरीज
Advertisements
उत्तर
\[\int\left( \frac{1 - \cos 2x}{1 + \cos 2x} \right)dx\]
`=∫ (2 sin^2 x) / ( 2 cos^2 x ) dx [∵ 1 - cos 2 θ = 2 sin^2 θ & 1 + cos 2 θ= 2 cos^2 θ]`
\[ = \int \tan^2 \text{x dx} \]
\[ = \int\left( \sec^2 x - 1 \right) dx\]
\[ = \int \sec^2\text{ x dx} - ∫ dx\]
\[ = \tan x - x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
` ∫ \sqrt{tan x} sec^4 x dx `
Evaluate the following integrals:
\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \text{ dx }\]
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
\[\int x \cos x\ dx\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int x^2 \text{ cos x dx }\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]
\[\int\frac{x}{\left( x - 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
