Advertisements
Advertisements
प्रश्न
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
बेरीज
Advertisements
उत्तर
\[\text{Let I} = \int\frac{\sec^2 x}{\ tanx + 2}dx\]
\[\text{Putting}\ \tan x = t\]
\[ \Rightarrow \sec^2 x = \frac{dt}{dx}\]
\[ \Rightarrow \sec^2 x dx = dt\]
\[ \therefore I = \int\frac{1}{t + 2}dt\]
\[ = \text{ln} \left| t + 2 \right| + C\]
\[ = \text{ln} \left| \ tan\ x + 2 \right| + C \left[ \because t = \tan x \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1}\text{ dx }\]
\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int x e^x \text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int x \sin x \cos x\ dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
\[\int\frac{1}{1 + \tan x} dx =\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{ dx }\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]
