Lutasin ang ∫ (from pi / 4 to pi / 2) of cosx/cosx wrt x

I-evaluate

\frac{\pi }{4}\approx 0.785398163

Mga Hakbang ng Solution

\int _ { \pi / 4 } ^ { \pi / 2 } \frac { \cos x } { \cos x } d x

I-divide ang \cos(x) gamit ang \cos(x) para makuha ang 1.

\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}1\mathrm{d}x

Suriin muna ang indefinite na integral.

\int 1\mathrm{d}x

Hanapin ang integral ng 1 gamit ang alituntunin ng talahanayan ng mga karaniwang integral \int a\mathrm{d}x=ax.

Ang definite integral ay ang antiderivative ng expression na na-evaluate sa upper limit ng integration minus ang antiderivative na na-evaluate sa lower limit ng integration.

\frac{\pi }{2}-\frac{\pi }{4}

Pasimplehin.

\frac{\pi }{4}

Quiz

Integration

5 mga problemang katulad ng:

\int _ { \pi / 4 } ^ { \pi / 2 } \frac { \cos x } { \cos x } d x

Katulad na mga Problema mula sa Web Search

How could I show that \int_{0}^{\frac{\pi}{2}} \frac{\cos (nx)}{\cos x}dx converges to zero as n\rightarrow \infty?

https://math.stackexchange.com/questions/1423757/how-could-i-show-that-int-0-frac-pi2-frac-cos-nx-cos-xdx-conv

This integral does not converge to 0 as n \to \infty. Case n = 2m. In this case, the integrand has simple pole at x = \frac{\pi}{2} \frac{\cos 2mx}{\cos x} \sim \frac{(-1)^m}{\frac{\pi}{2} - x} \quad \text{as } x \to \tfrac{\pi}{2}, ...

Integrating \int_0^{\pi/2} \cos^a(x) \cos(bx) \ dx

https://math.stackexchange.com/questions/456899/integrating-int-0-pi-2-cosax-cosbx-dx

\begin{align} &\int_0^{\frac\pi2}\cos^a(x)\cos(bx)\,\mathrm{d}x\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(e^{ix}+e^{-ix}\right)^ae^{ibx}\,\mathrm{d}x\right)\tag{1}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(1+e^{-2ix}\right)^ae^{i(a+b)x}\,\mathrm{d}x\right)\tag{2}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(1+e^{-2ix}\right)^ae^{i(a+b+2)x}\,\frac{i}{2}\mathrm{d}e^{-2ix}\right)\tag{3}\\ &=\mathrm{Re}\left(2^{-a}\int_1^{-1}\left(1+u\right)^au^{-(a+b)/2-1}\,\frac{i}{2}\mathrm{d}u\right)\tag{4}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{-1}\left(1+u\right)^au^{-(a+b)/2-1}\,\frac{i}{2}\mathrm{d}u\right)\tag{5}\\ &=\mathrm{Re}\left(2^{-a}\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}e^{i\pi((a+b)/2+1)}\,\frac1{2i}\mathrm{d}u\right)\tag{6}\\ &=\mathrm{Re}\left(2^{-a-1}e^{i\pi((a+b+1)/2)}\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}\,\mathrm{d}u\right)\tag{7}\\ &=2^{-a-1}\cos(\pi(a+b+1)/2)\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}\,\mathrm{d}u\tag{8}\\ &=2^{-a-1}\sin(-\pi(a+b)/2)\frac{\Gamma(a+1)\Gamma(-(a+b)/2)}{\Gamma(1+(a-b)/2)}\tag{9}\\ &=\frac{\pi\,2^{-a-1}\,\Gamma(a+1)}{\Gamma(1+(a-b)/2)\,\Gamma(1+(a+b)/2)}\tag{10} \end{align} ...

What is \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4x dx?

https://www.quora.com/What-is-int-limits_-frac-pi-2-frac-pi-2-cos-4x-dx

I will take a longer detour to find the answer to this question but it is worth the effort. Let’s denote J this integral. First, by symmetry, J=2\int_{0}^{\frac{\pi}{2}}\cos^{4}xdx Now, let ...

How will you integrate \displaystyle\int_{-\pi/2}^{\pi/2}\cos^{8}x \: \mathrm{d}x ?

https://www.quora.com/How-will-you-integrate-displaystyle-int_-pi-2-pi-2-cos-8-x-mathrm-d-x

I would use the fact that \cos(2x)=2\cos^2x-1, aka \cos^2x=\frac{\cos(2x)+1}{2}, multiple times. Use that to rewrite your cos^8, then you’ll have to expand the fourth power of a binomial, ...

Evaluating \int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx

https://math.stackexchange.com/q/2400460

Set \cos 2x = \sin^2\theta. Then \cos x = \sqrt{\frac{1+\sin^2\theta}{2}} \qquad \text{and} \qquad \left|\frac{dx}{d\theta}\right| = \frac{\sin\theta}{\sqrt{1+\sin^2\theta}}. So we have \int_{0}^{\frac{\pi}{4}} \cos^{3/2}(2x) \cos x \, dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{3\pi}{16\sqrt{2}}. ...

is this trig integral doable using contour integration?.

https://math.stackexchange.com/questions/381346/is-this-trig-integral-doable-using-contour-integration

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