Advertisements
Advertisements
प्रश्न
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
विकल्प
\[\frac{1}{2}\] log (sec x2 + tan x2) + C
\[\frac{x^2}{2}\] log (sec x2 + tan x2) + C
2 log (sec x2 + tan x2) + C
none of these
MCQ
Advertisements
उत्तर
\[\frac{1}{2}\] log (sec x2 + tan x2) + C
\[\text{ Let I }= \int x \sec x^2 dx\]
\[\text{ Putting x}^2 = t\]
\[ \Rightarrow 2x \text{ dx }= dt\]
\[ \Rightarrow x \text{ dx} = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\sec t \cdot dt\]
\[ = \frac{1}{2} \text{ log } \left| \sec t + \tan t \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \sec x^2 + \tan x^2 \right| + C \left( \because t = x^2 \right)\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int \tan^2 \left( 2x - 3 \right) dx\]
\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]
\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int x^2 e^{- x} \text{ dx }\]
` ∫ x tan ^2 x dx
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int\left( x + 1 \right) \sqrt{2 x^2 + 3} \text{ dx}\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{1}{1 + x + x^2 + x^3} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
