Question Statement

Evaluate the definite integral:

${\int }_0^{\frac{\pi }{4}}{\tan }^{-1}ydy$

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Background and Explanation

This problem requires integration by parts, a technique based on the product rule for differentiation. Since ${\tan }^{-1}y$ doesn't have an elementary antiderivative that is immediately obvious, we treat this as a product of ${\tan }^{-1}y$ and $1$, choosing $u={\tan }^{-1}y$ to simplify through differentiation.

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Solution

We begin by rewriting the integrand to explicitly show the product form suitable for integration by parts:

${\int }_0^{\frac{\pi }{4}}{\tan }^{-1}ydy={\int }_0^{\pi /4}{\tan }^{-1}y\cdot 1dy$

Applying integration by parts using the formula $\int udv=uv-\int vdu$:

Choose:

• $u={\tan }^{-1}y\Rightarrow du=\frac{1}{1+y^2}dy$
• $dv=1dy\Rightarrow v=y$

Substituting into the integration by parts formula:

$={\left[{\tan }^{-1}y\int 1dy-\int \left(\int 1dy\right)\frac{d}{dy}\left({\tan }^{-1}y\right)dy\right]}_0^{\pi /4}$

$={\left[{\tan }^{-1}y\cdot y-\int y\cdot \frac{1}{1+y^2}dy\right]}_0^{\pi /4}$

$={\left[y{\tan }^{-1}y\right]}_0^{\pi /4}-{\int }_0^{\pi /4}\frac{y}{1+y^2}dy$

Evaluating the first term and simplifying the remaining integral:

$=\left(\frac{\pi }{4}{\tan }^{-1}\frac{\pi }{4}-0\cdot {\tan }^{-1}(0)\right)-\frac{1}{2}{\int }_0^{\pi /4}\frac{2y}{1+y^2}dy$

$=\left(\frac{\pi }{4}(1)-0\right)-\frac{1}{2}{\left[\ln \left(1+y^2\right)\right]}_0^{\pi /4}$

Evaluating the definite integral:

$=\frac{\pi }{4}-\frac{1}{2}\left[\ln \left(1+{\left(\frac{\pi }{4}\right)}^2\right)-\ln (1+0)\right]$

$=\frac{\pi }{4}-\frac{1}{2}\left[\ln \left(1+\frac{{\pi }^2}{16}\right)-\ln 1\right]$

$=\frac{\pi }{4}-\frac{1}{2}\left[\ln \left(\frac{16+{\pi }^2}{16}\right)-0\right]$

$=\frac{\pi }{4}-\frac{1}{2}\ln \left(\frac{16+{\pi }^2}{16}\right)$

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Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• Derivative of Inverse Tangent: $\frac{d}{dy}({\tan }^{-1}y)=\frac{1}{1+y^2}$
• Logarithmic Integration: $\int \frac{f^{\prime }(y)}{f(y)}dy=\ln |f(y)|+C$ (specifically, recognizing $\frac{2y}{1+y^2}$ as the derivative of $\ln (1+y^2)$)
• Logarithm Properties: $\ln (a)-\ln (b)=\ln \left(\frac{a}{b}\right)$ and $\ln (1)=0$

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Summary of Steps

1. Rewrite the integral as ${\int }_0^{\pi /4}{\tan }^{-1}y\cdot 1dy$ to prepare for integration by parts
2. Choose $u={\tan }^{-1}y$ (to differentiate) and $dv=dy$ (to integrate), giving $v=y$ and $du=\frac{1}{1+y^2}dy$
3. Apply the integration by parts formula: ${\left[y{\tan }^{-1}y\right]}_0^{\pi /4}-{\int }_0^{\pi /4}\frac{y}{1+y^2}dy$
4. Evaluate the boundary term: $\frac{\pi }{4}{\tan }^{-1}\left(\frac{\pi }{4}\right)-0$
5. For the remaining integral, multiply and divide by 2 to get $\frac{1}{2}\int \frac{2y}{1+y^2}dy=\frac{1}{2}\ln (1+y^2)$
6. Evaluate the definite integral from $0$ to $\frac{\pi }{4}$: $\frac{1}{2}\left[\ln \left(1+\frac{{\pi }^2}{16}\right)-\ln (1)\right]$
7. Combine terms and simplify using logarithm properties to get the final result: $\frac{\pi }{4}-\frac{1}{2}\ln \left(\frac{16+{\pi }^2}{16}\right)$