Evaluate the integral. ∫cos−6xdx

Question

Evaluate the integral.   ∫cos−6xdx

Bukvald5z · Accepted Answer

Step 1  To evaluate the integral, ∫cos−6xdx  Solution:  The given integral can be written as,  ∫cos−6xdx=∫sec6xdx  Further solving integral as ,  ∫sec4xsec2xdx=∫(sec2x)2sec2xdx  =∫(1+tan2x)2sec2xdx  Step 2  Now substitute   t=tan⁡x  dt=sec2xdx  The integrate further becomes,  ∫(1+tan2x)2sec2xdx=∫(1+t2)2dt  =∫(1+t4+2t2)dt  =t+t55+2t33+c  substitute t as tanx and we get,  Step 3  ∫sec4xsec2xdx=tan⁡x+tan5x5+2tan3x3+c  Hence, the value of integral is tan⁡x+tan5x5+2tan3x3+c.

braodagxj · Accepted Answer

∫1cos6(x)dx  Rewrite / simplify using the definition of trigonometric / hyperbolic functions:  =∫sec6(x)dx  Prepare for replacement:  =∫sec2(x)(tan2(x)+1)2dx  u=tan⁡(x)⇒dudx=sec2x  =∫(u2+1)2du  We use the distributive property:  =∫(u4+2u2+1)du  Lets

nick1337 · Accepted Answer

Since R(−sin⁡(x)),−cos⁡(x))=R(sin⁡(x),cos⁡(x)), we make a trigonometric substitution: tan⁡(x)=t and then x=arctan⁡(t) dx=dt1+t2,sin⁡(x)=t1+t2,cos⁡(x)=11+t2 ∫(t2+1)2∗dt Simplify the expression : ∫(x2+1)2∗dx Calculate the tabular integral: ∫(x2+1)2∗dx=x55+2∗x33+x Answer: x55+2∗x33+x+C Returning to the change of variables (t=tan⁡(x)), we get: I=tan⁡(x)55+2∗tan⁡(x)33+tan⁡(x)+C

Evaluate the integral. \int \cos^{-6}xdx

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