12)
apply this property
\int_{-a}^{a}f(x)dx= \int_{0}^{a}(f(x)+f(-x))dx
Now
\int_{-\pi }^{\pi }{(cospx-sinqx)} ^{2} dx = \int_{0}^{\pi }({(cospx-sinqx)}^{2} + (cospx+sinqx)^{2})dx = 2 \int_{0}^{\pi }({(cospx)}^{2} + (sinqx)^{2})dx = 2(\pi)
8 Answers
11Thanks...
Can u pls give some hint for the first one.
i m getting no use of applying (a-x)property and so i added the old and new things but still the expression comes difficult...
any suggestions???
11)
first do
\int_{0}^{\pi /4}cot(\pi /4-x)ln(cos(\pi /4+x))dx
(using log property)
Now,
apply (a-x) property
\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx
so, u get
\int_{0}^{\pi /4} cot(\pi /4-\pi /4+x)ln(cos(\pi /4+\pi /4-x))dx = \int_{0}^{\pi /4}cotx lnsinx dx
now on evaluating this ans is coming not defined!!!
pls tell if i hve done any mistake!
62
\\\int_{0}^{\pi/4}{ln(cos(\pi/4+x))^{cot(\pi/4-x)}}dx \\=\int_{0}^{\pi/4}cot(\pi/4-x)\times{ln(cos(\pi/4+x))}dx \\=\int_{0}^{\pi/4}tan(\pi/4+x)\times{ln(cos(\pi/4+x))}dx
now substitute cos(pi/4+x) = t
\\=\int_{0}^{1/\sqrt{2}}1/t\times{ln(t)}dt
Which comes out to be undefined...
