Advertisements
Advertisements
प्रश्न
`int 1/(sin x - sqrt3 cos x) dx`
बेरीज
उत्तर
Given I = `int 1/(sin x - sqrt3 cos x) dx`
Let 1 = r cos θ and √3 = r sin θ
r = `sqrt(3 + 1) = 2`
And tan θ = √3 → θ = `pi/3`
=> `int 1/(sin x - sqrt3 cos x) dx = int 1/(rcos theta sin x - r sin theta cos x) dx`
= `1/r int 1/(sin (x - theta))dx`
= `1/r int cosec(x - theta)dx`
We know that `int cosec x dx = log|tan (x/2 - pi/6)| + c`
`1/2 log |tan(x/2 - pi/6)| + c`
∴ I = `int 1/(sinx - sqrt3 cos x) dx`
`1/2 log |tan (x/2 - pi/6)| + c`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
Integrate the following integrals:
\[\int\text { sin x cos 2x sin 3x dx}\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
` ∫ tan x sec^4 x dx `
\[\int\frac{1}{a^2 - b^2 x^2} dx\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]
\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int \tan^5 x\ dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\frac{x + 1}{x^2 + 4x + 5} \text{ dx}\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
