Let J=∫−5−4 3−x2tan⁡3−x2dx andK=∫−2−1 6−6x+x2tan⁡6x−x2−6dx. Then (J+K) equals __.

Let J=543x2tan3x2dx and

K=2166x+x2tan6xx26dx. Then (J+K) equals __.

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    Solution:

    We have J=543x2tan3x2dx

    Put (x+5)=t. Then

    J=013(t5)2tan3(t5)2dt=0122+10tt2tan22+10tt2dt

    Now K=2166x+x2tan6xx26dx.

    Put (x+2)=z. Then

    K=0166(z2)+(z2)2tan6(z2)(z2)26dz=012210z+z2tan22+10zz2dz

    Hence, (J+K)=0

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