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
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
If f(x) is a continuous function defined on [−a, a], then prove that
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int x^3/(x + 1)` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
