Question

)Find the product using identities
a)(x+4)(x+4)

Answer 1

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log(1 + tanx)dx}⟶I=∫04πlog(1+tanx)dx

Let above expression be equation (1).

We know that :-

\boxed{\rm \bf \displaystyle\int ^{ a} _{0} \bf f(x) \: dx= \bf \displaystyle\int ^{ a} _{0} \bf f(a - x) \: dx}∫0af(x)dx=∫0af(a−x)dx

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log(1 + tan( \frac{\pi}{4} - x))dx}⟶I=∫04πlog(1+tan(4π−x))dx

We know :-

\boxed { \bf tan(a - b) = \frac{tan \: a - tan \: b}{1 + (tan \: a \times tan \: b)} }tan(a−b)=1+(tana×tanb)tana−tanb

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg(1 + \frac{tan \frac{\pi}{4} - tan \: x}{1 + (tan\frac{\pi}{4} \times tan \: x)} \bigg)dx}⟶I=∫04πlog(1+1+(tan4π×tanx)tan4π−tanx)dx

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg(1 + \frac{1- tan \: x}{1 + tan \: x} \bigg)dx}⟶I=∫04πlog(1+1+tanx1−tanx)dx

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg( \frac{1 + tan \: x + 1- tan \: x}{1 + tan \: x} \bigg)dx}⟶I=∫04πlog(1+tanx1+tanx+1−tanx)dx

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg( \frac{2}{1 + tan \: x} \bigg)dx}⟶I=∫04πlog(1+tanx2)dx

We know :-

\boxed {\rm log( \dfrac{a}{b} ) = log \: a - log \: b }log(ba)=loga−logb

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm \bigg( log( 2) \: - log(1 + tan \: x) \bigg) dx⟶I=∫04π(log(2)−log(1+tanx))dx

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) \: dx \: - \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log(1 + tan \: x) dx⟶I=∫04πlog(2)dx−∫04πlog(1+tanx)dx

From equation (1) :-

\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) dx\: - I⟶I=∫04πlog(2)dx−I

\rm\longrightarrow \rm 2I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) \: dx⟶2I=∫04πlog(2)dx

\rm\longrightarrow \rm 2I = \displaystyle \rm log( 2)\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm1 \: dx⟶2I=log(2)

Answer 2

$\huge \underline \frak \pink {★彡A᭄ɴsᴡᴇʀ彡★}$

• x² + 16 + 8x