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int_(0)^(pi//8)2^(cos 4t)sin 4tdt

$\int_{0}^{\pi / 8} 2^{\cos 4 t} \sin 4 t d t$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{0}^{\pi / 8} 2^{\cos 4 t} \sin 4 t d t

Integrate with respect to u: Now we need to integrate $\frac{1}{4} \cdot 2^u$ with respect to $u$ . The integral of $2^u$ with respect to $u$ is $\frac{2^u}{\ln 2}$ , so we have: $\newline$ $\int 2^u \left(\frac{1}{4}\right) du = \frac{1}{4} \int 2^u du = \frac{1}{4} \cdot \frac{2^u}{\ln 2} + C$ $\newline$ We can now evaluate this antiderivative from $0$ to $1$ : $\newline$ $\left. \frac{1}{4} \cdot \frac{2^u}{\ln 2} \right|_{0}^{1} = \frac{1}{4 \ln 2} \cdot (2^1 - 2^0)$ $\newline$ $= \frac{1}{4 \ln 2} \cdot (2 - 1)$ $\newline$ $= \frac{1}{4 \ln 2}$
Evaluate antiderivative: We have now found the value of the definite integral: $\newline$ $\int_{0}^{\pi/8} 2^{\cos 4t} \sin 4t \, dt = \frac{1}{4 \ln 2}$

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