3 replies to this topic

[caybutbixanh]

Bài toán : Tính $\int _0^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+\sin 2x}}dx$

[chanhquocnghiem]

Bài toán : Tính $\int _0^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+\sin 2x}}dx$

$I=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+\sin2x}}\ dx=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+2\sin x\cos x}}\ dx=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{4-(\sin x-\cos x)^2}}\ dx$

Đặt $u=\sin x-\cos x\Rightarrow du=(\cos x+\sin x)dx$

$\Rightarrow I=\int_{-1}^{0}\frac{du}{\sqrt{4-u^2}}=\left [ \arcsin\frac{u}{2} \right ]_{-1}^{0}=\frac{\pi}{6}$.

[caybutbixanh]

$I=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+\sin2x}}\ dx=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{3+2\sin x\cos x}}\ dx=\int_{0}^{\frac{\pi}{4}}\frac{\cos x+\sin x}{\sqrt{4-(\sin x-\cos x)^2}}\ dx$

Đặt $u=\sin x-\cos x\Rightarrow du=(\cos x+\sin x)dx$

$\Rightarrow I=\int_{-1}^{0}\frac{du}{\sqrt{4-u^2}}=\left [ \arcsin\frac{u}{2} \right ]_{-1}^{0}=\frac{\pi}{6}$.

Bài 2: Tính $\int_0^{\frac{\pi}{4}} x.tan^2xdx$

Bài 3:Tính $\int_0^{\frac{\pi}{3}} \frac{x+\sin x}{\cos^2x}dx$

Bài 4: Tính $\int_0^{\frac{\pi}{2}} \cos ^2x.\cos ^2 2x dx$

[chanhquocnghiem]

Bài 2: Tính $\int_0^{\frac{\pi}{4}} x.tan^2xdx$

Bài 3:Tính $\int_0^{\frac{\pi}{3}} \frac{x+\sin x}{\cos^2x}dx$

Bài 4: Tính $\int_0^{\frac{\pi}{2}} \cos ^2x.\cos ^2 2x dx$

Bài 2 :

$I=\int_{0}^{\frac{\pi}{4}}x\tan^2xdx=\int_{0}^{\frac{\pi}{4}}x\left ( \frac{1}{\cos^2x}-1 \right )dx=\int_{0}^{\frac{\pi}{4}}\frac{xdx}{\cos^2x}-\int_{0}^{\frac{\pi}{4}}xdx$

Đặt $u=x$ ; $dv=\frac{dx}{\cos^2x}\Rightarrow v=\tan x$ ta có :

$I=\left [ x\tan x \right ]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\tan xdx-\int_{0}^{\frac{\pi}{4}}xdx=\left [ x\tan x+\ln\left | \cos x \right |-\frac{x^2}{2} \right ]_{0}^{\frac{\pi}{4}}=\frac{8\pi-\pi ^2-16\ln 2}{32}$.

 

Bài 3 :

$I=\int_{0}^{\frac{\pi}{3}}\frac{xdx}{\cos^2x}+\int_{0}^{\frac{\pi}{3}}\frac{\sin xdx}{\cos^2x}=\left [ x\tan x+\ln\left | \cos x \right |-\frac{1}{\cos x} \right ]_{0}^{\frac{\pi}{3}}=\frac{\pi \sqrt{3}-3\ln 2-3}{3}$.

 

Bài 4 :

$\int \cos^2x.\cos^22xdx=\int\frac{1+\cos2x}{2}.\cos^22xdx=\frac{1}{2}\int\cos^22xdx+\frac{1}{2}\cos^32xdx$

$=\frac{1}{2}\int\frac{1+\cos4x}{2}\ dx+\frac{1}{2}\int(1-\sin^22x)\cos2xdx$

$=\frac{1}{4}\int dx+\frac{1}{4}\int\cos4xdx+\frac{1}{2}\int\cos2xdx-\frac{1}{2}\int\sin^22x.\frac{1}{2}\ d(\sin2x)$

$=\frac{x}{4}+\frac{\sin4x}{16}+\frac{\sin2x}{4}-\frac{\sin^32x}{12}+C$

Vậy $I=\left [ \frac{x}{4}+\frac{\sin4x}{16}+\frac{\sin2x}{4}-\frac{\sin^32x}{12} \right ]_{0}^{\frac{\pi}{2}}=\frac{\pi}{8}$.

Edited by chanhquocnghiem, 10-03-2016 - 14:41.