Ví dụ tính tích phân hàm số lượng giác có lời giải

Công thức tính tích phân hàm lượng giác hay dùng:

$\begin{array}{l} \left( 1 \right)\,\,\,\,\,\,\,\,\int {\cos xdx}  = \sin x + C;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1′} \right)\,\,\,\,\,\,\,\,\int {\cos \left( {ax + b} \right)dx}  = \frac{1}{a}\sin \left( {ax + b} \right) + C;\,\,\\ \left( 2 \right)\,\,\,\,\,\,\,\,\int {\sin xdx}  =  – \cos x + C;\,\,\,\,\,\,\,\,\,\,\,\,\left( {2′} \right)\,\,\,\,\,\,\,\,\int {\sin \left( {ax + b} \right)dx}  =  – \frac{1}{a}\cos \left( {ax + b} \right) + C;\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\,\,\,\,\,\,\,\,\int {\frac{{dx}}{{{{\cos }^2}x}}}  = \int {\left( {1 + {{\tan }^2}x} \right)dx}  = \tan x + C;\,\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\,\,\,\,\,\,\,\,\int {\frac{{dx}}{{{{\sin }^2}x}}}  = \int {\left( {1 + {{\cot }^2}x} \right)dx}  =  – \cot x + C.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}$

*Chú ý: Các em cần phải nhớ các công thức lượng giác đã học.

Ví dụ 1.

$I\,\,\, = \,\,\,\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} \,\,\,\, = \,\,\,\,\frac{1}{2}\left. {\left( {\sin 2x} \right)} \right|_0^{\frac{\pi }{4}}\,\,\,\, = \,\,\,\,\frac{1}{2}.$

Ví dụ 2.

$I\,\,\, = \int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{\sin x + \cos x}}}  = \,\int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)}}} \,\,\, = \frac{1}{{\sqrt 2 }}\int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{4}} \right)}}}$ .

Đặt $t = x + \frac{\pi }{4} \Rightarrow dt = dx$. Đổi cận

$\begin{array}{c|lc}
x & 0 & \frac{\pi}{4} \\
\hline t & \frac{\pi}{4} & \frac{\pi}{2}
\end{array}$

Suy ra

$I\,\,\, = \frac{1}{{\sqrt 2 }}\int\limits_0^{\frac{\pi }{2}} {\frac{{dt}}{{{\mathop{\rm sint}\nolimits} }} = \frac{1}{{\sqrt 2 }}\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin tdt}}{{{{\sin }^2}t}} = } } \frac{1}{{\sqrt 2 }}\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin tdt}}{{1 – {{\cos }^2}t}}}$ .

Đổi cận

$\begin{array}{c|lr}
x & 0 & 2 \\
\hline t & -2 & 4 \\
\end{array}$

Suy ra

$I\,\,\, = \frac{1}{{\sqrt 2 }}\int\limits_{\frac{{\sqrt 2 }}{2}}^0 {\frac{{ – du}}{{1 – {u^2}}}}  = \frac{1}{{\sqrt 2 }}\int\limits_{\frac{{\sqrt 2 }}{2}}^0 {\frac{{du}}{{{u^2} – 1}}}  = \frac{1}{{2\sqrt 2 }}\left. {\left( {\ln \left| {\frac{{u – 1}}{{u + 1}}} \right|} \right)} \right|_{\frac{{\sqrt 2 }}{2}}^0 = \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{1 + \sqrt 2 }}{{1 – \sqrt 2 }}} \right|$.

Ví dụ 3.

$I\,\,\, = \int\limits_0^{\frac{\pi }{3}} {\sin 2x\cos xdx}  = \frac{1}{2}\int\limits_0^{\frac{\pi }{3}} {\left( {\sin 3x + \sin x} \right)dx}  = \frac{1}{2}\left. {\left( { – \frac{{\cos 3x}}{3} – \cos x} \right)} \right|_0^{\frac{\pi }{3}} = \frac{7}{{12}}$.

Ví dụ 4. Tính tích phân $I\,\,\, = \int\limits_0^{\frac{\pi }{2}} {\sin 2x{{\left( {1 + {{\sin }^2}x} \right)}^3}dx}$ .

Giải. Đặt $t = 1 + {\sin ^2}x \Rightarrow dt = 2\sin x\cos xdx \Rightarrow dt = \sin 2xdx$.

Đổi cận

$$\begin{array}{c|lr}
x & 0 & \frac{\pi}{2} \\
\hline t & 1 & 2
\end{array}$$

Suy ra

$I\,\,\, = \int\limits_1^2 {{t^3}dt}  = \left. {\left( {\frac{{{t^4}}}{4}} \right)} \right|_1^2 = \frac{{15}}{4}$.

Ví dụ 5. Tính tích phân $I\,\,\, = \,\,\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}x}}{{1 + \cos x}}dx}$ .

Giải. Đặt $t = 1 + \cos x \Rightarrow dt =  – \sin xdx$. Đổi cận

$\begin{array}{c|lr}
x & 0 & \frac{\pi}{2} \\
\hline t & 2 & 1
\end{array}$

Suy ra

$\begin{array}{l} I\,\,\, = \,\,\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^3}x}}{{1 + \cos x}}dx = } \int\limits_0^{\frac{\pi }{2}} {\frac{{1 – {{\cos }^2}x}}{{1 + \cos x}}\sin xdx} \,\, = \int\limits_2^1 {\frac{{1 – {{\left( {t – 1} \right)}^2}}}{t}\left( { – tdt} \right)} \\ \,\,\,\,\, = \int\limits_2^1 {\frac{{1 – {{\left( {t – 1} \right)}^2}}}{t}\left( { – tdt} \right)} \,\, = \int\limits_1^2 {\left( {2t – {t^2}} \right)dt}  = \left. {\left( {{t^2} – \frac{{{t^3}}}{3}} \right)} \right|_1^2 = \frac{2}{3}. \end{array}$

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