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1.[TEX]\int\limits_{0}^{\frac{\pi}{4}} \frac{dx}{cos^3x.cos( x - \frac{\pi}{4})}[/TEX]

2.[TEX]\int\limits_{0}^{\frac{\pi}{4}} ln(1 + tanx)dx[/TEX]

3.[TEX]\int\limits_{0}^{\frac{\pi}{4}} cos2x.ln(sinx + cosx)dx[/TEX]

4.[TEX]\int\limits_{0}^{\pi} \frac{x + sinx}{e^x}dx[/TEX]

5.[TEX]\int\limits_{0}^{e^{\pi}}[sin(lnx) + cos(lnx)]dx[/TEX]

6.[TEX]\int\limits_{0}^{\pi} e^{3x}(sinx + cosx)dx[/TEX]

 

[nghgh97]

1

\[\begin{array}{l}
\cos \left( {x - \frac{\pi }{4}} \right) = \frac{{\cos x + \sin x}}{{\sqrt 2 }}\\
I = \int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{{{\cos }^3}x.\cos \left( {x - \frac{\pi }{4}} \right)}}} = \sqrt 2 \int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{{{\cos }^3}x.\left( {\cos x + \sin x} \right)}}} \\
= \sqrt 2 \int\limits_0^{\frac{\pi }{4}} {\frac{{\left( {\cos x - \sin x} \right)dx}}{{{{\cos }^3}x.\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}} = \sqrt 2 \int\limits_0^{\frac{\pi }{4}} {\frac{{\left( {\cos x - \sin x} \right)dx}}{{{{\cos }^3}x.\left( {2{{\cos }^2}x - 1} \right)}}} \\
= \sqrt 2 \left( {\int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{{{\cos }^2}x.\left( {2{{\cos }^2}x - 1} \right)}}} - \int\limits_0^{\frac{\pi }{4}} {\frac{{\tan xdx}}{{{{\cos }^2}x.\left( {2{{\cos }^2}x - 1} \right)}}} } \right)
\end{array}\]
Bạn làm tiếp nhé.

 

[miko_tinhnghich_dangyeu]

$I=\int_0^{\dfrac{\pi}{4}} ln(1+tanx)dx$
Đặt $ x=\dfrac{\pi}{4}-t \Longrightarrow dx=-dt$
$I=\int_0^{\dfrac{\pi}{4}}ln(1+tan(\dfrac{\pi}{4}-t))dt=\int_0^{\dfrac{\pi}{4}}ln(\dfrac{2cost}{sint+cost})dt=\int_0^{\dfrac{\pi}{4}}ln2dt-\int_0^{\dfrac{\pi}{4}}ln(\dfrac{sint+cost}{cost})dt=\int_0^{\dfrac{\pi}{4}}ln2dt-\int_0^{\dfrac{\pi}{4}}ln(1+tant)dt$
vậy $I=\dfrac{1}{2}\int_0^{\dfrac{\pi}{4}}ln2dt$

 

[jet_nguyen]

Câu 3:
Ta có:
$$\int ^{\frac{\pi}{4}}_0 \cos2x\ln(\sin x+\cos x)dx$$$$=\int ^{\frac{\pi}{4}}_0 (\cos x-\sin x)(\cos x+\sin x)\ln(\sin x+\cos x)dx$$$$=\int ^{\frac{\pi}{4}}_0 (\cos x+\sin x)\ln(\sin x+\cos x)d(\cos x+\sin x)$$$$=\dfrac{1}{2}\int ^{\frac{\pi}{4}}_0 \ln(\sin x+\cos x)d(\cos x+\sin x)^2$$$$=\dfrac{(\cos x+\sin x)^2 \ln(\sin x+\cos x)}{2}\bigg|^{\frac{\pi}{4}}_0 -\dfrac{1}{2}\int ^{\frac{\pi}{4}}_0 (\sin x+\cos x)d(\sin x + \cos x)$$