[TEX]I= \int \frac{cos^2x}{sinx+ \sqrt{3}cosx}dx[/TEX]
Cách 1:
[TEX]I= \int \frac{cos^2x}{2sin(x+ \frac{ \pi}{3})}dx[/TEX]
Đặt [TEX]t=x+ \frac{ \pi}{3} \Rightarrow dx=dt, cosx= \frac{1}{2}(cost+ \sqrt{3}sint)[/TEX]
[TEX]\Rightarrow I= \frac{1}{8} \int \frac{(cost+ \sqrt{3}sint)^2}{sint}dt[/TEX]
[TEX]= \frac{1}{8} \int \frac{1+2 \sqrt{3}sintcost+2sin^2t}{sint}dt[/TEX]
[TEX]= \frac{1}{8} \int (2 \sqrt{3}cost++2sint)dt+ \frac{1}{8} \int \frac{1}{sint}dt[/TEX]
[TEX]= \frac{1}{8}(2 \sqrt{3}sint-2cost)- \frac{1}{8} \int [ \frac{1}{2(1-cost)}+ \frac{1}{2(1+cost)}]d(cost)[/TEX]
[TEX]= \frac{1}{4}( \sqrt{3}sint-cost)- \frac{1}{16}ln| \frac{1+cost}{1-cost}|+C[/TEX]
Cách 2:
[TEX] -3I= \int \frac{(sin^2x-3cos^2x)-sin^2x}{sinx+ \sqrt{3}cosx}dx[/TEX]
[TEX]= \int (sinx- \sqrt{3}cosx- \frac{1-cos^2x}{sinx+ \sqrt{3}cosx})dx[/TEX]
[TEX]=-cosx- \sqrt{3}sinx- \int \frac{dx}{sinx+ \sqrt{3}cosx}+I[/TEX]
[TEX]\Leftrightarrow 4I=cosx+ \sqrt{3}sinx+ \int \frac{dx}{sinx+ \sqrt{3}cosx}[/TEX]
[TEX]I_1= \int \frac{dx}{sinx+ \sqrt{3}cosx}= \frac{1}{2} \int \frac{dx}{sin(x+ \frac{ \pi}{3})}=- \frac{1}{4}ln| \frac{1+cos(x+ \frac{ \pi}{3})}{1-cos(x+ \frac{ \pi}{3})}|[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
