# Evaluate the integrals \int_0^{\pi/4}[\sec t i+\tan^2 t j-t\sin t k]dt

Evaluate the integrals
$$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{\sec{{t}}}{i}+{{\tan}^{{2}}{t}}{j}-{t}{\sin{{t}}}{k}\right]}{\left.{d}{t}\right.}$$

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Khribechy
To find the integral $$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{\sec{{t}}}{i}+{{\tan}^{{2}}{t}}{j}-{t}{\sin{{t}}}{k}\right]}{\left.{d}{t}\right.}$$
Explanation:
$$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{\sec{{t}}}{i}+{{\tan}^{{2}}{t}}{j}-{t}{\sin{{t}}}{k}\right]}{\left.{d}{t}\right.}={\left[{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\sec{{t}}}{\left.{d}{t}\right.}\right]}{i}+{\left[{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}^{{2}}{t}}{\left.{d}{t}\right.}\right]}{j}-{\left[{\int_{{0}}^{{\frac{\pi}{{4}}}}}{t}{\sin{{t}}}{\left.{d}{t}\right.}\right]}{k}$$
$$\displaystyle={I}_{{1}}{i}+{I}_{{2}}{j}-{I}_{{3}}{k}$$
Where $$\displaystyle{I}_{{1}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\sec{{t}}}{\left.{d}{t}\right.},{I}_{{2}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}},{I}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\sin{{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle{I}_{{1}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\sec{{t}}}{\left.{d}{t}\right.}$$
On integrating
$$\displaystyle={{\left[{\ln}{\left|{\sec{{t}}}+{\tan{{t}}}\right|}\right]}_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}$$
$$\displaystyle={\left[{\ln}{\left|{\sec{{\left({\frac{{\pi}}{{{4}}}}\right)}}}+{\tan{{\left({\frac{{\pi}}{{{4}}}}\right)}}}\right|}-{\ln}{\left|{\sec{{\left({0}\right)}}}+{\tan{{\left({0}\right)}}}\right|}\right]}$$
$$\displaystyle={\left[{\ln}{\left|\sqrt{{{2}}}+{1}\right|}-{\ln}{\left|{1}\right|}\right]}$$
$$\displaystyle={\left[{\ln}{\left|\sqrt{{{2}}}+{1}\right|}\right]}$$
$$\displaystyle{I}_{{2}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{{\tan}^{{2}}{t}}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\left({{\sec}^{{2}}{t}}-{1}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{{\sec}^{{2}}{t}}{\left.{d}{t}\right.}-{\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\left[{1}-{0}\right]}-{\left[{\frac{{\pi}}{{{4}}}}\right]}$$
$$\displaystyle={1}-{\frac{{\pi}}{{{4}}}}$$
$$\displaystyle{I}_{{3}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{t}{\sin{{t}}}{\left.{d}{t}\right.}$$
Aplying integration by parts
Recall the formula for integration by parts
$$\displaystyle\int{u}\cdot{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left[{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}\int{v}{\left.{d}{x}\right.}\right]}{\left.{d}{x}\right.}$$
Let $$\displaystyle{u}={t},{v}={\sin{{t}}}$$ and dx=dt
Therefore,
$$\displaystyle\int{t}{\sin{{t}}}{\left.{d}{t}\right.}={t}\int{\sin{{t}}}{\left.{d}{t}\right.}-\int{\left[{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}\right)}\int{\sin{{t}}}{\left.{d}{t}\right.}\right]}{\left.{d}{t}\right.}$$
$$\displaystyle={t}{\left(-{\cos{{t}}}\right)}-\int{\left[{1}\cdot{\left(-{\cos{{t}}}\right)}\right]}{\left.{d}{t}\right.}$$
$$\displaystyle={t}{\left(-{\cos{{t}}}\right)}+\int{\cos{{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle=-{t}{\cos{{t}}}+{\sin{{t}}}$$
$$\displaystyle\therefore{I}_{{3}}={\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{t}{\sin{{t}}}{\left.{d}{t}\right.}={{\left[-{t}{\cos{{t}}}+{\sin{{t}}}\right]}_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}$$
$$\displaystyle={\left[-{\frac{{\pi}}{{{4}}}}{\left({\frac{{{1}}}{{\sqrt{{{2}}}}}}\right)}+{\frac{{{1}}}{{\sqrt{{{2}}}}}}-{0}\right]}$$
$$\displaystyle={\frac{{{4}-\pi}}{{{4}\sqrt{{{2}}}}}}$$
Substituting $$\displaystyle{I}_{{1}},{I}_{{2}},{I}_{{3}}$$ value in equation (1), we get
$$\displaystyle{\int_{{0}}^{{{\frac{{\pi}}{{{4}}}}}}}{\left({\sec{{i}}}+{{\tan}^{{2}}{t}}{j}-{t}{\sin{{t}}}{k}\right)}{\left.{d}{t}\right.}={\ln}{\left|\sqrt{{{2}}}+{1}\right|}{i}+{\left[{1}-{\frac{{\pi}}{{{4}}}}\right]}{j}-{\left[{\frac{{{4}-\pi}}{{{4}\sqrt{{{2}}}}}}\right]}{k}$$
Answer: $$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{\sec{{t}}}{i}+{{\tan}^{{2}}{t}}{j}-{t}{\sin{{t}}}{k}\right]}{\left.{d}{t}\right.}={\ln}{\left|\sqrt{{{2}}}+{1}\right|}{i}+{\left[{1}-{\frac{{\pi}}{{{4}}}}\right]}{j}-{\left[{\frac{{{4}-\pi}}{{{4}\sqrt{{{2}}}}}}\right]}{k}$$