CBSE Class 12-science Answered

Solve it pls.

Asked by sdmbotch1123 | 30 Sep, 2023, 22:54: PM

Expert Answer

Integration can be solved only if

begin mathsize 14px style I space equals space integral fraction numerator square root of tan space x end root space over denominator cos to the power of 4 x end fraction space d x end style

Above integration is solved by rewriting above expression as follows

begin mathsize 14px style I space equals space integral square root of tan space x end root space s e c squared x space space s e c squared x space d x end style

......................(1)

let us use the substitution u = tan(x)

then, we have, du = sec² (x) dx

sec² (x) = 1 + tan² (x) = 1 + u²

hence expression (1) becomes

begin mathsize 14px style I space equals space integral space square root of u space left parenthesis space 1 space plus space u squared space right parenthesis space d u space space equals space integral space u to the power of 3 divided by 2 end exponent space d u space plus space integral space u to the power of 5 divided by 2 end exponent space d u end style

begin mathsize 14px style I space equals space 2 over 5 u to the power of 5 divided by 2 end exponent space plus space 2 over 7 space u to the power of 7 divided by 2 end exponent space space equals space 2 space u to the power of 5 divided by 2 end exponent space open parentheses 1 fifth space plus space 1 over 7 space u close parentheses end style

By resubstituting u = tanx , we get

begin mathsize 14px style I space equals space 2 space left parenthesis square root of tan space x end root space right parenthesis to the power of 5 space open parentheses 1 fifth space plus space 1 over 7 tan x close parentheses end style

Answered by Thiyagarajan K | 01 Oct, 2023, 08:58: AM

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