\int _{ }^{ }\sin ^2x\cos ^2x

1) Integrate the following functions:f(x)= (sinx+cosx)^2

What is the integral of the function f(x) = sin 2x?

Integrate the following functions:f(x)=sin⁡(5x)cos⁡(5x)

this case, a substitution method seems more straightforward. Let's use the substitution \(u = \sin(x)\), which implies that \(du = \cos(x) dx\). We can then express \(\cos^2(x)\) as \(1 - \sin^2(x)\) or \(1 - u^2\) to deal with the even power of cosine. The integral becomes: \[ \int u^3 (1 - u^2)^2 \cos(x) \, dx\] Since \(du = \cos(x) dx\), we can replace \(\cos(x) dx\) with \(du\): \[ \int u^3 (1 - u^2)^2 \, du\] Now we expand \((1 - u^2)^2\) and integrate term by term: \[ \int u^3 (1 - 2u^2 + u^4) \, du = \int (u^3 - 2u^5 + u^7) \, du\] Integrating each term separately: \[ \int u^3 \, du - 2 \int u^5 \, du + \int u^7 \, du = \frac{u^4}{4} - 2 \cdot \frac{u^6}{6} + \frac{u^8}{8} \] Simplify the coefficients: \[ \frac{u^4}{4} - \frac{u^6}{3} + \frac{u^8}{8} \] Now we substitute back \(u = \sin(x)\): \[ \frac{\sin^4(x)}{4} - \frac{\sin^6(x)}{3} + \frac{\sin^8(x)}{8} + C\] where \(C\) is the constant of integration. This is the antiderivative of \(\sin^3 x \cos^4 x\).

Find the integration of sin x cox(sin x)