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int((sin 2x)/(cos^(4)2x)+(1)/((2x+1)ln(2x+1)))dx=

$\int\left(\frac{\sin 2 x}{\cos ^{4} 2 x}+\frac{1}{(2 x+1) \ln (2 x+1)}\right) d x=$

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Evaluate definite integrals using the power rule

Full solution

Q.

\int\left(\frac{\sin 2 x}{\cos ^{4} 2 x}+\frac{1}{(2 x+1) \ln (2 x+1)}\right) d x=

Split Integral: First, let's split the integral into two parts: $\newline$ \int\left(\frac{\sin \(2\)x}{\cos^{\(4\)}\(2\)x} + \frac{\(1\)}{(\(2\)x+\(1\))\ln(\(2\)x+\(1\))}\right)dx = \int\frac{\sin \(2\)x}{\cos^{\(4\)}\(2\)x}dx + \int\frac{\(1\)}{(\(2\)x+\(1\))\ln(\(2\)x+\(1\))}dx
First Integral: Now, let's tackle the first integral \(\int\frac{\sin 2x}{\cos^{4}2x}\,dx. We can use a substitution: let $u = \cos(2x)$ , then $du = -2\sin(2x)\,dx$ . So, $-\frac{1}{2} du = \sin(2x)\,dx$ .
Substitution for First Integral: Substitute into the integral: $\newline$ $\int\frac{\sin 2x}{\cos^{4}2x}\,dx = \int\frac{-\frac{1}{2} du}{u^{4}} = -\frac{1}{2} \int u^{-4}\,du$
Integration of First Integral: Now, integrate $u^{-4}$ with respect to $u$ : $-\frac{1}{2} \int u^{-4}\,du = -\frac{1}{2} \cdot \left(-\frac{1}{3}\right)u^{-3} = \frac{1}{6}u^{-3}$
Replace $u$ in First Integral: Replace $u$ with $\cos(2x)$ to get back to $x$ : $\frac{1}{6}u^{-3} = \frac{1}{6}(\cos(2x))^{-3}$
Second Integral: Now, let's look at the second integral $\int \frac{1}{(2x+1)\ln(2x+1)}\,dx$ . This one is tricky, but we can use another substitution: let $v = \ln(2x+1)$ , then $dv = \frac{1}{(2x+1)}dx$ .
Substitution for Second Integral: Substitute into the integral: $\newline$ $\int \frac{1}{(2x+1)\ln(2x+1)}\,dx = \int \frac{1}{v} \,dv$
Integration of Second Integral: Now, integrate $1/v$ with respect to $v$ : $\int \frac{1}{v} \, dv = \ln|v|$
Replace $v$ in Second Integral: Replace $v$ with $\ln(2x+1)$ to get back to $x$ : $\ln|v| = \ln|\ln(2x+1)|$
Combine Results: Combine the results from both integrals: \int\left(\frac{\sin \(2\)x}{\cos^{\(4\)}\(2\)x} + \frac{\(1\)}{(\(2\)x+\(1\))\ln(\(2\)x+\(1\))}\right)dx = \frac{\(1\)}{\(6\)}(\cos(\(2x))^{ $-3$ } + \ln|\ln( $2$ x+ $1$ )| + C

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