Advertisements
Advertisements
Question
Solution
\[\text{Let I} = \int\frac{\cos 2x}{\left( \text{cos x }+ \text{sin x} \right)^2}dx\]
\[ = \int\frac{\cos^2 x - \sin^2 x}{\left( \text{cos x }+ \text{sin x} \right)^2}dx\]
\[ = \int\frac{\cos x - \ sin x}{\cos x + \sin x}dx\]
\[\text{Putting }\cos x + \sin x = t \]
\[ \Rightarrow - \text{sin x} + \text{cos x} = \frac{dt}{dx}\]
\[ \Rightarrow \left( \text{cos x}- \text{sin x} \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln }\left| t \right| + C\]
`= In | cos x + sin x |` + C ` [ ∵ t= cos x + sin x] `
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^3dx/(9+x^2)`
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)