Advertisements
Advertisements
प्रश्न
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
योग
Advertisements
उत्तर
Put `cos^2x` = t ⇒ `−2cosxsinxdx` = dt ⇒ `sin2xdx = -dt`
The given integral = `- int (dt)/sqrt(3^2 - t^2) = - sin^(-1) t/3 + c = - sin^(-1) (cos^2x)/3 + c`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
\[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 3 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]
\[\int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
