Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{Let I} = \int\frac{\cos 2x}{\left( \text{cos x }+ \text{sin x} \right)^2}dx\]
\[ = \int\frac{\cos^2 x - \sin^2 x}{\left( \text{cos x }+ \text{sin x} \right)^2}dx\]
\[ = \int\frac{\cos x - \ sin x}{\cos x + \sin x}dx\]
\[\text{Putting }\cos x + \sin x = t \]
\[ \Rightarrow - \text{sin x} + \text{cos x} = \frac{dt}{dx}\]
\[ \Rightarrow \left( \text{cos x}- \text{sin x} \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln }\left| t \right| + C\]
`= In | cos x + sin x |` + C ` [ ∵ t= cos x + sin x] `
APPEARS IN
संबंधित प्रश्न
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate:
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
`∫ (1)/(sin^2 x cos^2 x) dx`
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
