Advertisements
Advertisements
Question
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
Options
2(sinx + xcosθ) + C
2(sinx – xcosθ) + C
2(sinx + 2xcosθ) + C
2(sinx – 2x cosθ) + C
Advertisements
Solution
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to 2(sinx + xcosθ) + C.
Explanation:
Let I = `int (cos2x - cos 2theta)/(cosx - costheta) "d"x`
= `int ((2cos^2x - 1 - 2 cos^2theta + 1))/(cosx - costheta) "d"x`
= `2int ((cosx + cos theta)(cosx - costheta))/((cosx - costheta)) "d"x`
= `2int(cos x + cos theta) "d"x`
= 2(sinx + xcosθ) + C
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`Γ (9/2)`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
