2006-11-08 05:20:14 · 3 answers · asked by fm02 2 in Science & Mathematics ➔ Mathematics
yes.
I would solve first the square root of 1+sin2x
since (sinx)^2+(cosx)^2=1,so we can write
1+sin2x=(sinx)^2+(cosx)^2+sin2x
=(sinx)^2+(cosx)^2+2sinx cosx ( because we have the formula sin2x=2sinx cosx )
=(sinx +cosx )^2
so sq. root of 1+sin2x= sinx +cosx
now integral of sinx+cosx is -cosx+sinx
2006-11-08 05:36:37 · answer #1 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
Even after 10 yrs of high school, I can solve this.
sqrt(1 + sin2x) = sqrt(sinx ^ 2 + cosx ^ 2 + 2sinxcosx)
= sqrt((sinx + cosx) ^ 2) = sinx + cosx
2006-11-08 17:00:34 · answer #2 · answered by manoj Ransing 3 · 0⤊ 0⤋
? (sin(2x)+a million) / (a million-sin(2x)) enable, u = 2x du = 2 dx du/2 = dx. a million/2 ? (sinu+a million) / (a million-sinu) Now multiply the two precise and backside via --> (a million+sinu). a million/2 ? [ (sinu+a million)(a million+sinu) ] / [ (a million-sinu)(a million+sinu) ] du a million/2 ? (sin^2u + 2sinu + a million) / (a million-sin^2u) du a million/2 ? (sin^2u + 2sinu + a million) / cos^2u du enable's seperate the fractons. a million/2 ? sin^2u / cos^2u du + a million/2 ? 2sinu / cos^2u du + a million/2 ? a million / cos^2u du a million/2 ? tan^2u du + 2/2 ? tanusecu du + a million/2 ? sec^2u du a million/2 ? sec^2 - a million du + secu + (a million/2)tanu a million/2 [ tanu - u ] + secu + (a million/2)tanu + C (a million/2)tanu - (a million/2)u + secu + (a million/2)tanu + C (a million/2)tan(2x) - x + sec(2x) + (a million/2)tan(2x) + C desire this replaced into of help
2016-10-21 11:52:20 · answer #3 · answered by Anonymous · 0⤊ 0⤋