Advertisements
Advertisements
प्रश्न
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
\[\int \tan^2 \left( 2x - 3 \right) dx\]
` ∫ tan x sec^4 x dx `
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int \sec^4 x\ dx\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
