Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
बेरीज
Advertisements
उत्तर
`int 1/[ 2 - 3cos2x] dx`
As `cos 2x = 2cos^x - 1`
So `int 1/[ 2 - 3cos2x] dx = int 1/[2 - 3( 2cos^x - 1) ]`
And multiply and divide by sec2x
Then we have `int sec^2x/[5sec^2x - 6]` dx
= `int (sec^2x)/[5( 1 + tan^2x) - 6]`
= `int (sec^2x dx)/( 5tan^2x - 1)`
Let tan x = t, then sec2x dx = dt
Hence `int [sec^2x]/[ 5tan^2x - 1]`
= `int dt/[5t^2 - 1]`
= `1/5 int dt/[t^2 - (1/sqrt5)^2]`
= `1/5 log |[ t - 1/sqrt5 ]/[ t + 1/sqrt5 ]|`
= `1/5 log |[ tan x - 1/sqrt5]/[ tan x + 1/sqrt5 ]| + c`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
` ∫ sec^6 x tan x dx `
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{1}{a^2 - b^2 x^2} dx\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\frac{1}{1 - \cot x} dx\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int x \cos x\ dx\]
\[\int x e^x \text{ dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int\log \left( x + \sqrt{x^2 + a^2} \right) \text{ dx}\]
