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
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f(2a − x) = −f(x), prove that
If f is an integrable function, show that
Evaluate each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
Find: `int logx/(1 + log x)^2 dx`
