ada yang bisa jawab soal integral?

4. misal
[tex]u=\frac{1}{x}[/tex], maka [tex]\frac{du}{dx}=-\frac{1}{x^2}\rightarrow \frac{1}{x^2} dx=-du[/tex]
[tex]\int{\frac{\sin{\frac{1}{x}}}{x^2} \, dx=\int{-\sin{u}} \,du\\=\cos{u}+C\\=\cos{\frac{1}{x}}+C[/tex]

1.
[tex]\sin{x+\frac{\pi}{6}}\cos{x-\frac{pi}{6}}=\frac{1}{2}\times\left(\sin{2x}+\sin{\frac{\pi}{3}}\right)\\ =\frac{1}{2}\times\left(\sin{2x}+\frac{\sqrt{3}}{2}\right)[/tex]
[tex]\int_{0}^{\frac{\pi}{6}}{\sin{\left(x+\frac{\pi}{6}\right)}\cos{\left(x-\frac{pi}{6}}\right)} \,dx=\int_{0}^{\frac{\pi}{6}}{\frac{1}{2}\times\left(\sin{2x}+\frac{\sqrt{3}}{2}\right)} \,dx[/tex]
[tex]=\frac{1}{2}\times \left[-\cos{2x}+\frac{\sqrt{3}}{2}x\right]_{0}^{\frac{\pi}{6}}[/tex]
[tex]=\frac{1}{2}\times \left[\left(-\cos{\left(2\times\frac{\pi}{6}\right)}+\frac{\sqrt{3}}{2}\times \frac{\pi}{6}\right)-\left(-\cos{0}+\frac{\sqrt{3}}{2}\times 0\right)\right][/tex]
[tex]=\frac{1}{2}\times\left[-\frac{1}{2}+\frac{\sqrt{3}\pi}{12}+1\right][/tex]
[tex]=\frac{1}{4}+\frac{\sqrt{3}\pi}{24}[/tex]

3. misalkan
[tex]u=2x^3-5[/tex], maka [tex]\frac{du}{dx}=6x^2\rightarrow 2x^2 dx=\frac{1}{3}du[/tex]
[tex]\int{\frac{2x^2}{\sqrt[7]{2x^3-5}} \,dx=\int{\frac{1}{3}\frac{1}{u^{\frac{1}{7}}} \,du[/tex]
[tex]=\int{\frac{1}{3}u^{-\frac{1}{7}}} \,du[/tex]
[tex]=\frac{1}{3}\frac{7}{6}u^{\frac{6}{7}}+C[/tex]
[tex]=\frac{7}{18}\left(2x^3-5\right)^{\frac{6}{7}}+C[/tex]

3.
[tex]2-4\sin^{2}{x}=2\left(1-2\sin^{2}{x}\right)=2\cos{2x}[/tex]
[tex]\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{2-4\sin^{2}{x}} \,dx=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{2\cos{2x}} \,dx[/tex]
[tex]=\left[\sin{2x}\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}[/tex]
[tex]=\sin{\left(2\times \frac{3\pi}{4}\right)}-\sin{\left(2\times \frac{\pi}{4}\right)}[/tex]
[tex]=(-1)-1[/tex]
[tex]=-2[/tex]