Advertisements
Advertisements
प्रश्न
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
योग
Advertisements
उत्तर
` Note:" Here, we are considering "log x as log_e x `
\[\text{Let I} = \int\frac{1 + \cot x}{x + \log \sin x}dx\]
\[\text{Putting}\ x + \log \ sin\ x = t\]
\[ \Rightarrow 1 + \ cot\ x = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 + \cot x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log} \left| t \right| + C\]
\[ = \text{log }\left| x + \log \sin\ x \right| + C \left[ \because t = x + \log \sin x \right]\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int \cos^2 \text{nx dx}\]
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
\[\int x^3 \sin x^4 dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{2x + 5}{x^2 - x - 2} \text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int2 x^3 e^{x^2} dx\]
\[\int x \cos^3 x\ dx\]
\[\int e^x \left( \cos x - \sin x \right) dx\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{1}{\sin x \left( 3 + 2 \cos x \right)} dx\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int \tan^3 x\ dx\]
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int x \sec^2 2x\ dx\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
