nguyenngocmai

Tìm nguyên hàm :$I=\int \frac{\sin 3x +\sin 2x}{\sqrt{2+ \cos x}}dx$

$I=\int \frac{3sinx-4sin^{3}x}{\sqrt{2+cosx}}dx+\int \frac{sin2x}{\sqrt{2+cosx}}dx$

$=\int \frac{-sinx}{\sqrt{2+cosx}}dx+\int \frac{4cos^{2}xsinx}{\sqrt{2+cosx}}dx+\int \frac{2sinxcosx}{\sqrt{2+cosx}}dx$

$=\int \frac{1}{\sqrt{2+cosx}}dcosx-\int \frac{4cos^{2}x}{\sqrt{2+cosx}}dcosx-\int \frac{2cosx}{\sqrt{2+cosx}}dcosx$

$=2\sqrt{2+cosx}-I_{1}-I_{2}$

$giải I_{1},I_{2} bằng cách đặt ẩn phụ t=\sqrt{2+cosx}$

$\int_{0}^{1}(x^{2}sinx^{3}+\frac{\sqrt{x}}{1+x})dx=\int_{0}^{1}x^{2}sinx^{3}dx+\int_{0}^{1}\frac{\sqrt{x}}{1+x}dx$

$\int_{0}^{1}x^{2}sinx^{3}dx=\int_{0}^{1}\frac{sinx^{3}}{3}dx^{3}$

$\int_{0}^{1}\frac{\sqrt{x}}{1+x}dx=I$

đặt t=$\sqrt{x}$

$\Rightarrow$ I=$\int_{0}^{1}\frac{2t^{2}}{1+t^{2}}dt$

đặt t=tank

$\int_{1}^{2}\frac{1}{x^{4}+2x}dx=\int_{1}^{2}\frac{1}{2}\left ( \frac{1}{x} -\frac{x^{2}}{x^{3}+2}\right )dx$