Question

sin³x + cos³x / sin cos²x
dx​

Answer 1

Answer:

$\huge\pink{elloh \: \: sis\: jii}$

Step-by-step explanation

Determine partial fractions  

(sin²x*cos²x) / (sin³x + cos³x) = (-3cos(x)^4 + 6 sin²(x) cos²(x) + 2cos²(x) - 3 sin(x)^4 + 2sin²(x)^2 + 1)/[6(cos(x)+sin(x)) (cos²(x)-2sin(x)cos(x) + sin²(x)+1)] (only need to know that cos²(x) = 1-sin²(x) for this)  

= (1 / ( sin(x) + cos(x) ) +4(sin(x)+cos(x))/((cos(x)-sin(x))²+1)- 3sin(x)-3cos(x))/6  

Step two:  

Integrate partial fractions:  

∫1 / ( sin(x) + cos(x) ) dx = √2 / 2 [ ln(1+ ( tan(x/2)-1) / √2) - ln (1-( tan(x/2)-1) / √2) ]  

∫ 4(sin(x)+cos(x))/((cos(x)-sin(x))²+1) dx = -4 * arc tan ( cos(x)-sin(x) )  

∫ -3sin(x)-3cos(x) dx = 3cos(x)-3sin(x)  

Step three:  

Determine the anti-derivative:  

1/6 ( -4 arc tan ( cos(x)-sin(x) ) + √2/ 2 [ ln ( 1+ ( tan(x/2)-1) / √2) - ln ( 1 - ( tan(x/2)-1) / √2) ] + 3 cos(x) – 3 sin(x) )  

Step four:  

Evaluate the integral at the interval boundaries:  

π/4  

∫ (sin²x*cos²x) / (sin³x + cos³x) dx =  

0  

= (π-3+√2/2 [ ln(1+1/√2) - ln(1-1/√2) ] + √2/2 ln ( 1+ ( tan(π/8)-1) / √2 ) - ln ( 1 - ( tan(π/8)-1) / √2 ) )/6