Question

Evaluate the following integrals...​

Answer 1

${\orange{ \boxed{\boxed{\begin{array}{cc}\displaystyle \int \rm \: \sqrt{1 + cos3x} \: \: dx \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \:Substitute \: \: \: u = 3x \\ \implies \rm \: du = 3 \: dx \\ \therefore \rm \: \frac{1}{3} du = \: dx \end{array}}}} \\ \\ = \frac{1}{3} \displaystyle \int \rm \: \sqrt{1 + cos \: u} \: \: du\\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \: we \: know \: that \\ \rm1 + cos2x = 2 {cos}^{2}x \\ \therefore \: 1 + cos \: x = 2 {cos}^{2} \frac{x}{2}\end{array}}}} \\ \\ = \frac{1}{3} \displaystyle \int \rm \: \sqrt{2{cos}^{2} \frac{u}{2} } \: \: du</p><p> \\ \\ = \frac{1}{3} \displaystyle \int \rm \: \sqrt{2} \: cos \frac{u}{2} \: \: du \\ \\ \pink{ {\boxed{\begin{array}{cc}\rm \:Substitute \: \: v = \frac{u}{2} \\ \implies \rm \: dv = \frac{1}{2} \: du \\ \therefore \rm \: du = 2 \: dv\end{array}}}} \\ \\ = \frac{1}{3} \displaystyle \int \rm \: \sqrt{2} \: cos \: v \: \: .2 \: dv \\ \\ = \frac{1}{3} \times {2}^{ \frac{3}{2} } \displaystyle \int \rm \: cos \: v \: \: dv \\ \\\pink{ {\boxed{\begin{array}{cc}\rm \:\displaystyle \int \rm \:cos \: x \: \: dx = sin \: x + c\end{array}}}} \\ \\ = \rm \: \frac{1}{3} \times {2}^{ \frac{3}{2} } \: sin \: v + c \\ \\ \rm = \frac{1}{3} \times {2}^{ \frac{3}{2} }.sin \frac{u}{2} + c \\ \\ \rm = \frac{1}{3} \times {2}^{ \frac{3}{2} } .sin \frac{3x}{2} + c \\ \\ \rm = \frac{ {2}^{ \frac{3}{2} } .sin \frac{3x}{2} }{3} + c \\ \\ \sf = \: answer\end{array}}}}}$

$\pink{\ddot{\smile}}$

Answer 2

$\large\underline{\bold{Given \:Question - }}$

Evaluate the following integral

$\displaystyle\int\sf \sqrt{1 + cos3x} \: dx$

$\red{\large\underline{\sf{Solution-}}}$

The given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \sqrt{1 + cos3x} \: dx$

We know,

$\boxed{ \bf{ \:1 + cos2x = {2cos}^{2}x}}$

So, using this identity, we get

$\rm \:  =  \:\displaystyle\int\sf \sqrt{2 {cos}^{2} \bigg[\dfrac{3x}{2} \bigg]} \: dx$

$\rm \:  =  \: \sqrt{2}\displaystyle\int\sf cos\bigg[\dfrac{3x}{2} \bigg] \: dx$

We know,

$\boxed{ \bf{ \:\displaystyle\int\sf cos(ax + b) \: dx \: = \: \frac{sin(ax + b)}{a} + c \: \: }}$

$\rm \:  =  \: \sqrt{2} \times \dfrac{sin\bigg[\dfrac{3x}{2} \bigg]}{\dfrac{3}{2} } + c$

$\rm \:  =  \: \dfrac{2 \sqrt{2} }{3} \: sin\dfrac{3x}{2} \: + \: c$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$