Question

integrate sin+cosx/sin^2x+cos^4x​

Answer 1

Given Question :-

Evaluate :-

$\displaystyle\int \tt \: \dfrac{sinx + cosx}{ {sin}^{2}x + {cos}^{4}x} \: dx$

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

We know that,

$\boxed{\bf \: {sin}^{2}x = 1 - {cos}^{2}x }$

$\boxed{\bf \: {cos}^{2}x = 1 - {sin}^{2}x }$

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ {a}^{2}-{x}^{2}} = \dfrac{1}{2a}log |\dfrac{a + x}{a - x}| + c}$

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ {x}^{2} + {a}^{2}} = \dfrac{1}{a} {tan}^{ - 1}\dfrac{x}{a} + c}$

Let's solve the problem now!!

Let we first represent the denominator in terms of sinx - cosx, so that derivative of this exist in numerator.

Thus,

Consider,

$\rm :\longmapsto\: {sin}^{2}x + {cos}^{4}x$

$\rm = \:\: 1 - {cos}^{2}x + {( {cos}^{2}x) }^{2}$

$\rm = \:\:1 - {cos}^{2}x(1 - {cos}^{2}x)$

$\rm = \:\:1 - {cos}^{2}x {sin}^{2}x$

$\rm :\implies\: {sin}^{2}x + {cos}^{4}x = 1 - {cos}^{2}x {sin}^{2}x - - - (1)$

Consider,

$\rm :\longmapsto\: {(sinx - cosx)}^{2}$

$\rm \: {(sinx - cosx)}^{2} = \:\: {sin}^{2}x + {cos}^{2}x - 2sinxcosx$

$\rm \: {(sinx - cosx)}^{2} = \:\: 1 - 2sinxcosx$

$\rm :\longmapsto\:2sinxcosx = 1 - {(sinx - cosx)}^{2}$

$\rm :\implies\:sinxcosx = \dfrac{1 - {(sinx - cosx)}^{2} }{2} - - (2)$

Substituting this value in equation (1), we get

$\rm \:{sin}^{2}x+{cos}^{4}x = 1 - {\bigg(\dfrac{1 - {(sinx - cosx)}^{2} }{2} \bigg) }^{2}$

Now,

Given Integral,

$\rm :\longmapsto\:Let \: I = \: \displaystyle\int \tt \: \dfrac{sinx + cosx}{ {sin}^{2}x + {cos}^{4}x} \: dx$

can be rewritten as

$\rm \: I= \:\:\displaystyle\int \tt \: \dfrac{sinx + cosx}{1 - {\bigg(\dfrac{1 - {(sinx - cosx)}^{2} }{2} \bigg) }^{2}} \: dx$

As we know that,

By using substitution method, we have

• ⇒ Put sinx - cosx = y

Differentiate w.r.t x, we get.

• ⇒ ( cosx + sinx ) dx = dy

So,

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

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{4dy}{4 - {(1 - {y}^{2})}^{2} }$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{4dy}{ {2}^{2} - {(1 - {y}^{2})}^{2} }$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{4dy}{(2 + 1 - {y}^{2})(2 - 1 + {y}^{2})}$

$\: \: \: \: \: \: \: \: \: \: \: \purple{\boxed{ \bf \because \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{4dy}{(3 - {y}^{2})(1+ {y}^{2})}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{(3 + 1)dy}{(3 - {y}^{2})(1+ {y}^{2})}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{(3 + 1 + {y}^{2} - {y}^{2} )dy}{(3 - {y}^{2})(1+ {y}^{2})}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{(1 +{y}^{2}) + (3 -{y}^{2})dy}{(3 - {y}^{2})(1+ {y}^{2})}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{(1 +{y}^{2})dy}{(3 - {y}^{2})(1+ {y}^{2})} + \displaystyle\int \tt \: \dfrac{(3 - {y}^{2})dy}{(3 - {y}^{2})(1+ {y}^{2})}$

$\rm \:I\:=\:\displaystyle\int \tt \: \dfrac{dy}{ {( \sqrt{3}) }^{2} - {y}^{2}} + \displaystyle\int \tt \: \dfrac{dy}{{y}^{2} + {1}^{2} }$

$\rm \:I\: = \: \dfrac{1}{2 \sqrt{3} }log |\dfrac{ \sqrt{3} + y}{ \sqrt{3} - y} | + {tan}^{ - 1}y + c$

$\rm \:I= \dfrac{1}{2 \sqrt{3} }log\bigg|\dfrac{\sqrt{3}+(sinx-cosx)}{ \sqrt{3} - (sinx - cosx)} \bigg|+{tan}^{-1}(sinx-cosx)+c$

$\rm \:I=\dfrac{1}{2 \sqrt{3} }log\bigg|\dfrac{\sqrt{3}+sinx-cosx}{ \sqrt{3} - sinx + cosx} \bigg|+{tan}^{-1}(sinx-cosx)+c$

Additional Information :-

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ {x}^{2}- {a}^{2}} = \dfrac{1}{2a}log |\dfrac{x - a}{x + a} | + c}$

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ \sqrt{{a}^{2} - {x}^{2}}}= {sin}^{ - 1} \dfrac{x}{a} + c}$

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ \sqrt{{x}^{2} - {a}^{2}}} =log |x + \sqrt{{x}^{2} - {a}^{2}} | + c}$

$\boxed{\bf \: \displaystyle\int \tt \: \dfrac{dx}{ \sqrt{{x}^{2} + {a}^{2}}} =log |x + \sqrt{{x}^{2} + {a}^{2}} | + c}$