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Solving: 2π∫0π43sin⁡(x)1+9cos⁡(x)2dx2\pi \int_{0}^{\frac{\pi }{4}}3\sin\left(x\right)\sqrt{1+9\cos\left(x\right)^2}dx
Solve instead: 2π ∫0π 4 3sin(x)1+9cos2(x)dy2\pi\:\int_0^{\frac{\pi\:}{4}}\:3sin\left(x\right)\sqrt{1+9cos^2\left(x\right)}dy
f(x)=3sin(x)(1+9cos(x)^2)^(1/2)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

2π ∫0π 4 3sin(x)1+9cos2(x)dy2\pi\:\int_0^{\frac{\pi\:}{4}}\:3sin\left(x\right)\sqrt{1+9cos^2\left(x\right)}dy

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Final answer to the exercise

−15.6292262−πln⁡(2.3452079+2.1213204)+9.424778⋅3.1622777−−πln⁡((1+9)0.5+3)-15.6292262-\pi \ln\left(2.3452079+2.1213204\right)+9.424778\cdot 3.1622777- -\pi \ln\left(\left(1+9\right)^{0.5}+3\right)

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Main Topic: Integrals of Rational Functions

Integrals of rational functions of the form R(x) = P(x)/Q(x).

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2π ∫0π 4  3sin(x)√1+9cos2(x)dy
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