#Trigonometric Integrals and Substitutions
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<@&286206848099549185>
In your textbook there should be a topic about signs of cos and sin
what to do when the exponents is odd, even, etc
It’s easy to make mistakes when you do intermediate steps in your head
yes
i fixed that mistake
but it gave me at the end pi/2
then i redid it again
i got pi/2
i cant get the right answer
ill send you my two solutions
<@&286206848099549185>
you forgot the bounds of integration
if you do u(pi) and u(0) you dont have to resubstitute the value of u after evaluating integral. if you keep same bounds you have to resubstitute. if you do double u substitution keep track like a book keeper or a russian egg doll
\begin{align}
\int_{0}^{\pi}cos^6xdx//
&= \frac{1}{8} \int_{0}^{2\pi}(1+cosu)^3du\
&= \frac{1}{16}\right(\int_{0}^{2\pi}du + 3\int_{0}^{2\pi}cosudu + \frac{3}&={2}\int_{0}^{2\pi} (1 + cos2u)du + 0 \left)\
&= \frac{1}{16} \left( u\bigl |{0}^{2\pi} + 3sinu \bigl |{0}^{2\pi} + \frac{3}{2}\left(u + \frac{1}{2}sin2u \right) \bigl |_{0}^{2\pi} \right)\
&= \frac{1}{16} (2\pi + \frac{3}{2}(2\pi)\
&= \frac{1}{16}(5\pi)
\end{align}
jassmeene
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im so confused why do i need to change the limits giveno
i had a bunch of substitution rule exercices i never did it and i got it all right
did you resubstitute the value of u at the end when evaluating the definite integral?
how many times did you do substitution?