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Why is $\int_0^{\pi/4} 5(1+\tan(x))^3\sec^2(x)\,dx$ equal to $18.75$ and not $3.75$?

I know the indefinite integral $\int 5(1+\tan(x))^3\sec^2(x)\,dx= \frac {5(1+\tan(x))^4} {4}+c$ by using $u$ substitution. Then shouldn't I evaluate that at $\frac{\pi}4$ minus that at $0$? Doing that gives $3.75$ but my texbook and wolfram alpha say the right answer is $18.75$

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  • $\begingroup$ It might be beneficial to us if you add your work so we can analyze what went wrong $\endgroup$ – Crescendo Dec 18 '17 at 23:15
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Note that we have $\tan\left(\frac{\pi}{4} \right)=1$ and $\tan(0)=0$

Hence,

\begin{align} \frac{5(1+\tan\left(\frac\pi4 \right))^4}{4}-\frac{5(1+0)^4}{4}&=\frac{5(2^4)}{4}-\frac{5}{4}\\&=\frac{5(15)}{4}\\&= \color{blue}5(3.75)\\&=18.75 \end{align}

A potential mistake is that you have forgotten to multiply by $5$. We have $\frac{15}4=3.75$ but I can't tell for sure unless more working is shown.

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