Advertisements
Advertisements
प्रश्न
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
पर्याय
2(sinx + xcosθ) + C
2(sinx – xcosθ) + C
2(sinx + 2xcosθ) + C
2(sinx – 2x cosθ) + C
Advertisements
उत्तर
`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
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Find: `int logx/(1 + log x)^2 dx`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
