The integral ∫1(t+1)2dt can be solved using the formula:∫x−ndx=x−n+1−n+1+CThus,∫1(t+1)2dt=−1t+1+C Step 6: Combine ResultsPutting everything together:I=x−(−1t+1)+C=x+1tanx+1+C Final ResultThus, the final result of the integral is:∫sin2x(sinx+cosx)...
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2、积分出来的结果不确定原因也有两个: 第一、由于有大量的代数、对数、三角的恒等式的出现, 可以在积分结果中互相替换;这种恒等式,在英文 中,用的的formulae。 第二、三角函数中有无数的恒等式,这个恒等式,在英文 中才是identity。3、举例如下:
cos(x2)+sin(x2)=8 解x (復數求解) x=(−1)((−i)ln(4+4i+(−21−21i)×6221)+2πn1)21,n1∈Z x=((−i)ln(4+4i+(−21−21i)×6221)+2πn1)21,n1∈Z x=(−1)((−i)ln(4+4i+(21+21i)×6221)+2...
∫f(x)dx=F(x)+c, where F'(x) = f(x). Required formulae to solve this problem: sin(2x)=2sinxcosxsin(3x)=3sinx−4sin3xcos(2x)=1−2sin2x∫cos(ax)dx=sin(ax)a+c∫sin(ax)dx=−cos(ax)a+c ...
Using the half-angle formula for cosine:cosx2=√1+cosx2Substituting cosx=−√154:cosx2= ⎷1−√1542= ⎷44−√1542=√4−√158=√4−√152 Step 4: Find tanx2Using the identity:tanx2=sinx2cosx2Substituting the values we found:tanx2=√4+√152√4−√152=√4+√15√4−√...
This can be rewritten using the sine addition formula sin(a+b)=sinacosb+cosasinb. Step 4: Identify coefficientsWe can identify:- 12=sinπ6- √34=cosπ6 Thus, we can rewrite the left side as:2(sin(x+π6))=2√2 Step 5: Simplify the equationNow we simplify the equation:sin(x+π...