Advertisements
Advertisements
Question
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
Sum
Advertisements
Solution
\[\text{Let I} = \int\frac{1 - \sin x}{x + \cos x}dx\]
\[\text{Putting x} + \cos x = t\]
\[ \Rightarrow 1 - \sin x = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 - \sin x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln t} + C\]
\[ = \text{ln }\left| x + \cos x \right| + C \left[ \because t = x + \cos x \right]\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \sin^5 x \text{ dx }\]
Evaluate the following integrals:
\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{2x}{2 + x - x^2} \text{ dx }\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int x \sin x \cos 2x\ dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int \sec^6 x\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
