Question

(b) 5. Identify the pattern and find the number of triangles in the nth figure. 会 11​

Answer 1

Answer:

It can be observed that in the given matchstick pattern, the number of matchsticks are 3.5,7, and 9, which is 1 more than twice Of the number Of triangles in the pattern.

Hence, the pattern is 2n+1, where n is the number of triangles.

Answer 2

Answer:

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

Given integral is

$\displaystyle\int\rm \sqrt{cotx} \: dx$

To evaluate this integral, we substitute

$\rm \: \sqrt{cotx} = y$

$\rm \: cotx = {y}^{2}$

$\rm \: - {cosec}^{2}x dx\: = \: 2y \: dy$

$\rm \: - (1 + {cot}^{2}x )dx\: = \: 2y \: dy$

$\rm \: - (1 + {y}^{4})dx\: = \: 2y \: dy$

$\rm \: dx \: = \: - \dfrac{2y}{ {y}^{4} + 1} \: dy$

On substituting all these values in given integral, we get

$\rm \: = - \: \displaystyle\int\rm y \times \frac{2y}{ {y}^{4} + 1} \: dy$

$\rm \: = - \: \displaystyle\int\rm \frac{2 {y}^{2} }{ {y}^{4} + 1} \: dy$

$\rm \: = - \: \displaystyle\int\rm \frac{{y}^{2} + {y}^{2} }{ {y}^{4} + 1} \: dy$

$\rm \: = - \: \displaystyle\int\rm \frac{{y}^{2} + {y}^{2} + 1 - 1}{ {y}^{4} + 1} \: dy$

$\rm \: = - \: \displaystyle\int\rm \frac{{y}^{2}+1}{ {y}^{4} + 1}dy -\displaystyle\int\rm \frac{{y}^{2} - 1}{ {y}^{4} + 1}dy$

On divide numerator and denominator by y^2 in both integrals, we have

$\rm \: = - \displaystyle\int\rm \frac{1 + \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} } }dy - \displaystyle\int\rm \frac{1 - \dfrac{1}{ {y}^{2} } }{ {y}^{2}+\dfrac{1}{ {y}^{2} } }dy$

$\rm \: = - \displaystyle\int\rm \frac{1 + \dfrac{1}{ {y}^{2} } }{ {\bigg(y - \dfrac{1}{y} \bigg) }^{2} + 2}dy - \displaystyle\int\rm \frac{1 - \dfrac{1}{ {y}^{2} } }{ {\bigg(y + \dfrac{1}{y} \bigg) }^{2} - 2 }dy$

Now, to evaluate further these integrals, we substitute

$\rm \: y - \dfrac{1}{y} = u \: \: and \: \: y + \dfrac{1}{y} = v$

$\rm \: \bigg(1 + \dfrac{1}{ {y}^{2} }\bigg)dy= du \: \: and \: \: \bigg(1 - \dfrac{1}{ {y}^{2} }\bigg)dy = dv$

So, on substituting these values in above integral, we get

$\rm \: = - \displaystyle\int\rm \frac{du}{ {u}^{2} + (\sqrt{2})^{2} } - \displaystyle\int\rm \frac{dv}{ {v}^{2} - {( \sqrt{2} )}^{2} }$

$\rm \: = - \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{u}{ \sqrt{2} } - \dfrac{1}{2 \sqrt{2} }log\bigg |\dfrac{v - \sqrt{2} }{v + \sqrt{2} } \bigg| + c$

$\rm \: = - \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{y - \dfrac{1}{y}}{ \sqrt{2} } - \dfrac{1}{2 \sqrt{2} }log\bigg |\dfrac{y + \dfrac{1}{y} - \sqrt{2} }{y + \dfrac{1}{y} + \sqrt{2} } \bigg| + c$

$\rm \: = - \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{ {y}^{2} - 1}{ \sqrt{2} y} - \dfrac{1}{2 \sqrt{2} }log\bigg |\dfrac{ {y}^{2} + 1 - \sqrt{2} y}{ {y}^{2} + 1 + \sqrt{2}y } \bigg| + c$

$\rm \: = - \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{cotx - 1}{ \sqrt{2} \sqrt{cotx} } - \dfrac{1}{2 \sqrt{2} }log\bigg |\dfrac{cotx + 1 - \sqrt{2} \sqrt{cotx} }{cotx + 1 + \sqrt{2} \sqrt{cotx} } \bigg| + c$

Hence,

$\color{green}\rm\implies \:\displaystyle\int\rm \sqrt{cotx} \: dx \\ \\ \color{green}\rm \: = - \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{cotx - 1}{ \sqrt{2} \sqrt{cotx} } - \dfrac{1}{2 \sqrt{2} }log\bigg |\dfrac{cotx + 1 - \sqrt{2} \sqrt{cotx} }{cotx + 1 + \sqrt{2} \sqrt{cotx} } \bigg| + c$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\rm \: \frac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

$\boxed{ \rm{ \:\rm \: \frac{d}{dx} k \: = \: 0 \: }}$

$\boxed{ \rm{ \:\rm \: \frac{d}{dx} cotx \: = \: - \: {cosec}^{2}x \: }}$

$\boxed{ \rm{ \: {cosec}^{2}x - {cot}^{2}x = 1 \: }}$

$\boxed{ \rm{ \: {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy = {(x - y)}^{2} + 2xy \: }}$

$\boxed{ \rm{ \:\displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c \: }}$

$\boxed{ \rm{ \:\displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} log\bigg | \frac{x - a}{x + a} \bigg| + c \: }}$