Question

Integration sin5 x.cos5 x dx

Answer 1

GivEn

To integrate : sin5xcos5x

SoluTion

Let us consider

$\sf \implies \sin5x = t \\ \\ \sf Differentiating \: both \: sides \: with \: respect \: to \: x\\ \\ \implies \sf \dfrac{d}{dx} ( \sin5x) = \frac{dt}{dx} \\ \\ \implies \sf \cos5x \times 5 = \dfrac{dt}{dx} \\ \\ \sf\implies 5\cos5x \: dx = dt$

Now on integration we have

$\sf \longrightarrow \int \sin5x \cos5x \: dx \\ \\ \longrightarrow \sf \dfrac{5}{5} \int\sin5x \cos5x \: dx \\ \\ \longrightarrow \sf \dfrac{1}{5} \int 5\sin5x \cos5x \: dx \\ \\ \longrightarrow \sf\dfrac{1}{5} \int \sin5x \times5 \cos5x \: dx \\ \\ \longrightarrow \sf\dfrac{1}{5}\int t \: dt \\ \\ \longrightarrow \sf \dfrac{1}{5} \{ \dfrac{t^{2}}{2} + C\} \\\\ \sf \longrightarrow \dfrac{\sin^{2}5x}{10} + C$

Answer 2

$\bold{\int sin5x.cos5x\;dx...(1)}$

Let u = sin5x ... (2)

$\implies \bold{du=5.cosxdx }\\\\\implies \bold{\frac{du}{5}=cosx.dx...(3)}$

Now sub. (2) & (3) in (1) , we get ,

$\implies \bold{\int u.\frac{du}{5}}\\\\\implies \bold{\frac{1}{5}\int u.du}\\\\\implies \bold{\frac{1}{5}.\frac{u^2}{2}+c}\\\\\implies \bold{\frac{sin^2(5x)}{10}+c \;[From\;(2)]}$

$\bold{\bf{\red{So\;,\;\int sin5x.cos5x.dx=\frac{sin^2(5x)}{10}+c }}}$