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

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 $18.75$

• It might be beneficial to us if you add your work so we can analyze what went wrong – Crescendo Dec 18 '17 at 23:15

Note that we have $\tan\left(\frac{\pi}{4} \right)=1$ and $\tan(0)=0$
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.