Đáp án:
Giải thích các bước giải:
I7)
\(\int {{{\tan }^4}xdx} = \int {{{\tan }^2}x.{{\tan }^2}xdx} \) \( = \int {\left( {{{\tan }^2}x - 1 + 1} \right){{\tan }^2}xdx} \) \( = \int {\left( {{{\tan }^2}x + 1} \right){{\tan }^2}xdx} - \int {{{\tan }^2}xdx} \)
Đặt \(t = \tan x\) \( \Rightarrow dt = \left( {1 + {{\tan }^2}x} \right)dx\)
\( \Rightarrow \int {\left( {{{\tan }^2}x + 1} \right){{\tan }^2}xdx} \) \( = \int {{t^2}dt} = \dfrac{{{t^3}}}{3} + C = \dfrac{{{{\tan }^3}x}}{3} + C\)
Lại có: \(\int {{{\tan }^2}xdx} = \int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}dx} \)
\( = \int {\dfrac{1}{{{{\cos }^2}x}}dx} - \int {dx} \) \( = \tan x - x + C\)
Vậy \(\int {{{\tan }^4}xdx} = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C\)
I8)
Đặt \(t = \sqrt {x + 1} \Rightarrow {t^2} = x + 1\) \( \Rightarrow 2tdt = dx\)
Khi đó \(\int {{e^{\sqrt {x + 1} }}dx} = \int {{e^t}.2tdt} \)\( = 2\int {{e^t}tdt} \)
Đặt \(\left\{ \begin{array}{l}u = t\\dv = {e^t}dt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right.\)
\( \Rightarrow 2\int {{e^t}tdt} = 2\left( {t{e^t} - \int {{e^t}dt} } \right)\) \( = 2\left( {t{e^t} - {e^t}} \right) + C\) \( = 2\left( {t - 1} \right){e^t} + C\) \( = 2\left( {\sqrt {x + 1} - 1} \right){e^{\sqrt {x + 1} }} + C\)
I10) Đặt \(\left\{ \begin{array}{l}u = \cos \left( {\ln x} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = - \dfrac{1}{x}\sin \left( {\ln x} \right)dx\\v = x\end{array} \right.\)
\( \Rightarrow \int {\cos \left( {\ln x} \right)dx} = x\cos \left( {\ln x} \right) + \int {\sin \left( {\ln x} \right)dx} \)
Đặt \(\left\{ \begin{array}{l}u = \sin \left( {\ln x} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{x}\cos \left( {\ln x} \right)dx\\v = x\end{array} \right.\)
\( \Rightarrow \int {\sin \left( {\ln x} \right)dx} = x\sin \left( {\ln x} \right) - \int {\cos \left( {\ln x} \right)dx} \)
Suy ra \(\int {\cos \left( {\ln x} \right)dx} = x\cos \left( {\ln x} \right) + x\sin \left( {\ln x} \right) - \int {\cos \left( {\ln x} \right)dx} + C\)
\( \Leftrightarrow 2\int {\cos \left( {\ln x} \right)dx} = x\left( {\cos \left( {\ln x} \right) + \sin \left( {\ln x} \right)} \right) + C\) \( \Leftrightarrow \int {\cos \left( {\ln x} \right)dx} = \dfrac{1}{2}\left[ {x\left( {\cos \left( {\ln x} \right) + \sin \left( {\ln x} \right)} \right) + C} \right]\)
Vậy \(\int {\cos \left( {\ln x} \right)dx} = \dfrac{1}{2}\left[ {x\left( {\cos \left( {\ln x} \right) + \sin \left( {\ln x} \right)} \right)} \right] + C\)
