Solve the problem in photo showing each step clearly 

Asked by Vidushi412 | 28th Aug, 2019, 09:41: PM

Expert Answer:

I equals integral fraction numerator square root of sin x end root d x over denominator square root of sin x end root plus square root of cos x end root end fraction
equals integral fraction numerator d x over denominator 1 plus square root of co t x end root end fraction
T a k e space square root of c o t x end root equals t rightwards double arrow fraction numerator 1 over denominator 2 square root of c o t x end root end fraction open parentheses negative cos e c squared x close parentheses space d x equals d t
d x equals fraction numerator negative 2 t space d t over denominator 1 plus t to the power of 4 end fraction
therefore space I equals integral fraction numerator negative 2 t space d t over denominator open parentheses 1 plus t to the power of 4 close parentheses open parentheses 1 plus t close parentheses end fraction
equals negative integral fraction numerator open parentheses 1 plus t to the power of 4 space plus 1 minus t to the power of 4 close parentheses t space d t over denominator open parentheses 1 plus t to the power of 4 close parentheses open parentheses 1 plus t close parentheses end fraction
equals negative open square brackets integral fraction numerator t over denominator 1 plus t end fraction d t plus integral fraction numerator open parentheses 1 plus t squared close parentheses open parentheses t minus t squared close parentheses over denominator 1 plus t to the power of 4 end fraction d t close square brackets
N o w space i n t e g r a t e space t h i s space a n d space g e t space t h e space r e s u l t

Answered by Renu Varma | 30th Aug, 2019, 10:51: AM