Answer complete please 
 

Asked by Varsneya Srinivas | 28th Dec, 2017, 12:29: PM

Expert Answer:

begin mathsize 16px style integral fraction numerator tan 2 straight theta over denominator square root of cos to the power of 6 straight theta plus sin to the power of 6 straight theta end root end fraction dθ integral fraction numerator tan begin display style 2 end style begin display style straight theta end style over denominator square root of begin display style open parentheses cos squared straight x plus sin squared straight x close parentheses cubed minus 3 cos squared xsin squared straight x open parentheses cos squared straight x plus sin squared straight x close parentheses end style end root end fraction dθ integral fraction numerator tan begin display style 2 end style begin display style straight theta end style over denominator square root of 1 minus 3 cos squared xsin squared straight x end root end fraction dθ integral fraction numerator tan begin display style 2 end style begin display style straight theta end style over denominator square root of 1 minus begin display style 3 over 4 end style open parentheses 4 sin squared θcos squared straight theta close parentheses end root end fraction dθ integral fraction numerator tan begin display style 2 end style begin display style straight theta end style over denominator square root of 1 minus begin display style 3 over 4 end style sin squared 2 straight theta end root end fraction integral fraction numerator tan begin display style 2 end style begin display style straight theta end style over denominator square root of 1 minus begin display style 3 over 4 end style open parentheses 1 minus cos squared 2 straight theta close parentheses end root end fraction dθ integral fraction numerator tan 2 straight theta over denominator square root of 1 minus begin display style 3 over 4 end style plus begin display style 3 over 4 end style cos squared 2 straight theta end root end fraction dθ integral fraction numerator tan 2 straight theta over denominator square root of fraction numerator 1 plus 3 cos squared 2 straight theta over denominator 4 end fraction end root end fraction dθ integral fraction numerator 2 sin 2 straight theta over denominator cos 2 straight theta square root of 1 plus 3 cos squared 2 straight theta end root end fraction dθ cos 2 straight theta equals straight t minus 2 sin 2 θdθ equals dt integral fraction numerator negative dt over denominator straight t square root of 1 plus 3 straight t squared end root end fraction integral fraction numerator negative tdt over denominator straight t squared square root of 1 plus 3 straight t squared end root end fraction 1 plus 3 straight t squared equals straight u 6 tdt equals du Now space integrate space further. end style

Answered by Sneha shidid | 28th Dec, 2017, 02:08: PM

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